All right, it is six o'clock. Since we have a quorum, we can go ahead and call the meeting to order. Not sure if everybody has the agenda in front of them, but are there any requests to make any changes or additions? The only agenda I've seen is the executive session. Is there another agenda? Yes, the full board agenda should have been, was destroyed by Amy Jewel a little week ago. So you would have received an email from Amy with the full packet. Perhaps you can, maybe Amy can resend that to John. Well, let me first. I'm sending now. Okay. Thank you. Before we just kind of get started, one thing that I've been thinking about with now that we're only six plus months into this, is that to the extent that your internet connection can handle it, we do always appreciate if you can have your video on. I know there are other circumstances that may make it so that that's not possible, but it does help with the community feeling and also just, I think there's a lot of reliance on nonverbal language of like head nodding. And so to the extent that you can and feel comfortable and don't have other things going on that are distractions, we certainly welcome that. So thank you. John, did you have your hand up for something? Not me. Okay. All right. Do we have any members of the public on the call? We do. Is there anybody who would like to comment at this time? No, it has raised their hand to comment. So we're good. Okay. I might need to turn my volume up so I can hear you. All right. So the consent agenda, are there any requests to remove items for discussion? Hearing none. We will approve that. Um, the next item on the agenda is the preliminary budget. Tara, if you'd like to. Yep. This is a requirement of our charter to have the board, basically the board proposes a preliminary budget and essentially what it's designed to do is to alert our member towns to any assessments that may be coming in the next year. So we need to get this out for charter. We're a little bit late and it's supposed to be out by December one. Nola has prepared a memo that she can walk through with you, but is this is not, it's certainly not the final budget. This is what we anticipate starting with going into the budget season. This is essentially a requirement of the charter. So go ahead, Nola. So this is typically based on the inflation rate, which you can see from the other information is 1.2%. So it's a pretty basic calculation across the board. We're just saying, okay, everything is gonna be 1.2% more. And I averaged that where I'm sorry, I took the higher value of whichever was either in the actuals or in the budget, just to be sure that we were covering all of our bases within this, but it looks like we're covered. We still have plenty of money in the stabilization reserve. And like Sarah said, this isn't our current or what our budget will be at the end of this fiscal year. Does it have anything to do with that? I move we accept the proposed budget. I guess do you quote an interim budget or? So you appropriate? Yes, we have, there's a proposed motion on the memo, but that's a very good question Leslie to raise. So one thing that I want the newer members to know is that this is something that Nola and I have talked about. I also discussed it with our most, not our most recent finance director, but the one previous where I think we need to take a look at the requirements in the charter. It does require the board to present a proposed budget. We are offering a preliminary estimate of the budget, but when really in reality, the charter wants us to have the next year's budget done by December 1. That has always been a struggle for the 25 years we've been in the pleasant existence, yet it is a requirement. And again, mainly it is so that our member communities can make any adjustments that they need to do for their own budgets ahead of town meeting day. So we are looking at some budgeting processes and budgeting software that we feel may be able to allow us to actually meet the requirements of the charter, not necessarily for next fiscal year, but for the year after. So that if we implement the software, it will be able to get that information in and then in use for fiscal 23. So Leslie, yes, your question is appropriate to ask because what the charter says is proposed, what we are doing here is preliminary. So this is again, just our first snapshot based on rate inflation and CPI, but it is not exactly what we, I think are meant to do for the charter. The other option is to change the charter and that is not an easy doing. So the more that we can adapt our own processes to the existing charter language, I think that should be our goal. So there is a proposed motion and it says, be it resolved that the board of commissioners acknowledges the receipt of your preliminary budget to be further developed through the normal budget process. And I would add that for fiscal year 2022. So moved. Essex seconds. The other item that is important to know is that we don't plan on dunning the towns or any money out of their budgets. And that's what they need to know for their March budget. So that's the reason for us to have this bottom line number that we know we're gonna be able to make using our own incomes. Correct. So that is the intent that we do not, we're not proposing that we will be issuing any assessments to any member of town in fiscal 22. We haven't had the executive session yet. Well, true, but we are confident we will not need to do that, even with that. And we'll tell you why in the executive session. Any further discussion? All right, all in favor? Aye. Aye. Aye. Any opposed? Any abstain? All right, motion passes. Thank you. Thank you very much. Thank you, Nova. And the next item is personnel staffing. Yes, so I would like to bring Amy and Josh Tyler into this conversation. We had, I believe preview this for the board last month to let you know that we were planning to bring this to you this month. And this is an area where we feel particularly vulnerable and that is in our maintenance staff and particularly our drivers. So I would like actually Josh, if you could kind of walk through some of the highlights of the memo and then Amy can discuss any of the financial justifications that are included if there are any questions there. So Josh. All right. Can everybody hear me? All right, great. So historically the maintenance and rollout department had five employees working for them. We had four kind of maintenance and rolloff workers separated between the maintenance and the rolloff programs.  After the departure of our most recent supervisor it kind of was apparent that it made more sense to kind of move that supervisory role into a project manager's position because it kind of that guidance and oversight was more appropriate to that position where the people who were doing the direct service that manager that supervisory role didn't really fit. And that was kind of an oversight piece on our part that we wanted to have a little more control. So due to COVID budgeting, we dropped that fifth position. We were gonna keep it in for potentially hiring a fifth in the event we needed a fifth person down there but we dropped it now under our existing conditions of people schedules with schools closing at a whim and just people needing to take vacations and just our kind of our increase in managing our own internal materials. It makes sense that we kind of add that fifth position back. We had planned on adding it back in FY22 anyway but just under our current situations it makes more sense to have a more robust department and that's doing that is adding that fifth position in. When Sarah says vulnerable, we only have four drivers right now. All our maintenance and roll-off positions have a CDL driver's license so that means they can manage our MSW, our recycling, our metal, things that they need to haul on a large scale. In the event that one of our, we lose one of our people to COVID it gets really tight really quick. We can get by with three, if we drop down to two it's pretty difficult to manage all of our facility needs and that would drive us to start looking at either reducing services at our facilities or reducing the actual number of facilities that are open. So we've kind of identified that this fifth position is necessary and it kind of gives us that flexibility in the event that one of our staff down at maintenance roll-off happens to get COVID. We don't have to go straight to looking at what service we can and can't provide or looking at what facility we need to close. So that's kind of a part of it and just the overall, we're running thin down there as it is. We're not, we're keeping up with our maintenance requirements but a lot of times our maintenance crew is pulled to do roll-off driving. So that's kind of the basis of this request. And again, to Josh's point, we've restructured the department so that this is truly, before we had a working supervisor, but we really needed a full-time working worker. So they will be kind of splitting time between also some maintenance duties, so some mechanic work. And that will also have another driver will free up the folks that we have on the team who are really good mechanics to do more of that work instead of doing the driving. So we're looking to play to the existing skills of the team and you'll bring in, again, that much needed security for the driver's position. So Amy did look at the wage benefit and chart that we have in the fiscal 21 budget. And we were really thrilled that the health insurance increase only came in just over 1% if it was one and a half percent, Amy. And instead of what we had budgeted, which was around 10%. So right there, there's significant savings that we won't have to shell out in January. So, Amy, I don't know if you want to fill in any additional info on the funding of the position. Sure, so when we create the budget and we start that in January, typically with the finance committee, we go through our wages and benefits. So all of the staffing and all of the benefits are laid out there. And then when it became apparent that we need another roll-off truck driver, maintenance operator position, we looked at that spreadsheet and it was great that we already had the MVP information in there as well. We also had budgeted for a project manager position, July one, and just hired that position. So there was a bit of savings available there. And then we've also had a couple of folks who have been out on leave. And so their salary has decreased. So looking at projecting from now through June, we knew that we had about $165,000 under budget within the wage bend. So we knew we could hire for this position if it was approved by the board, we have money there. And that of course assumes that there are any drastic changes and benefits that we don't have single folks who go to family plans given our structure, but that typically has not changed and open enrollment just happened. We didn't have any major changes. So we're pretty comfortable that that number is accurate. And this is an entry-level position. This is just straight-up maintenance and roll-off. It's not mechanic, it's just to carry, share the load. Literally. Yeah. Exactly. Any questions from the commissioners? Bren, I have a couple of questions. It seems to me the presentation indicates that there's enough savings in the current year's budget to cover the position, but two questions. Is there an estimate for the cost of the new position through the balance of this current fiscal year? And the second question would be what have you identified any savings from contracted services? Because that's identified in this justification memo. So I think both of those dollar figures would be helpful to know. I can help you with that. All in at an entry-level position with benefits is roughly $60,000 to $70,000. We did a quick look on wage-ben to see what that was. The market's a little tight right now, but that's kind of our expectation of where we're gonna land. The savings we'll see with this extra person is, we've got a couple of really good mechanics right now, but if they have to be on the road, then some of the equipment that we have, we have to send to out to a shop to get some routine maintenance done or some things that we can do potentially in-house. So there's that savings there. And on that same vein, when we send stuff to a shop, say a loader or even a roll-off truck or a service truck, we have to wait for the shop to get to it. So it can sit there up to a week, which could potentially limit some of our maintenance needs. So there will be some savings in there. It also allows us to say looking forward, potentially internally haul more of our own materials because we do contract some of those out right now. And we're taking a look at a few of them, particularly organics. It's looking like we're gonna spend about $100,000 this year managing organics internally, managing organics through a third-party vendor. And with this fifth position, it gives us the flexibility to potentially manage that ourselves. So that'll be a significant savings. It'll be a shift in operations and it will carry savings. We have to look at the capital expense to do that. But over a five, 10-year period, it's a significant savings. And Josh, is that what you're looking for? It's 60 to 70,000 is annual salary. Annual, correct. That's it for clarifying, yep. Other questions? All right, does anyone? There is a resolution that I can read. Be it resolved that the board of commissioners authorizes the executive director to hire for the position of maintenance operator, roll-off truck driver. I'll move that. Mr. John from Milton. Friendly amendment. I think it should say entry-level position. I'm fine with that. Okay. Is there any issues with that, Sarah? No. Do I hear a second? Second. Logan Richmond. Any further discussion? I'm just checking, is this required or is this a more of a courtesy notification that's? It's technically a change to the budget because we did not include the position in the budget even at a zero that we wanted to, I felt that we needed to get the permission of the board to alter the budget. Good. All in favor? Hi. Hi. Any opposed? Any abstain? Motion passes. Thank you all very much. Thank you. All right. Next, the swift. So that is Josh Estee. So Josh can pop on and use. Hi all, can everybody hear me? Yes. Apologies for looking like I'm working by candlelight. I'm not, I'm pulled up in a spare bedroom. So as the memo states, Kasella reached out to interested parties at the end of November and requested that the new B transfer station be re-included in our SWIP. If folks remember or those who were on the board, they came to us at the end of last year and requested based on an amendment they were sending to ANR to re-include their transfer station because they're adding their de-packager operations. So this second amendment to their facility management plan amends that further to reflect a change of operations for their de-pack operations, they have added some fixed onsite tank storage to their transfer station. Those are up already. And they're also looking to amend the total volume of organics that they're actually moving through the transfer station. So in looking at the red line strikeout version of their facility management plan, those were the two big items. I did see Brynn, you emailed me that I hadn't attached the red line strikeout version of the facility management plan. I'll be happy to share that if anybody were interested. It's a number of pages long, so it would take a little bit to scroll through, but I could provide the highlights unless you want me to just provide it after the meeting. One point I'd like to make, this is sort of the first step as a reminder for ANR to start the amendment process. So they need to check the box that the district that the facility is in has approved them in their sweat before they will go through the amendment process. I spoke with Dennis Feckert with ANR who indicated to me that it will be a major amendment. And all that really means is that they lengthen the period of open comment period to about 30 days. So this is, their red line strikeout of their facility management plan is something that we can and potentially will comment on. This, the approval by the board here tonight doesn't disallow us of having that opportunity. So we'll be meeting as a staff probably in the next week or so to just discuss the changes that they're proposing and talk about whether or not we'd like to comment on them or not. Okay, I have a question. Go for it. Yes. If I understand what we're doing correctly, it's really not our facility, but it's their facility. We have to approve it because we're kind of like governing body over with. Is there any act 250 implication here? Sounds like they're gonna be changing their capacity. And when they applied, I'm sure act 250 had a limit of what they were doing. If this changes that, do they need to or do we need to make sure act 250 is covered? That's a good question. As part of it, we will follow up and make sure that they will because if you read our facility inclusion plan, it talks about making sure that they are following all state and local permits. So that will be part of our due diligence. We generally like to go to someone in the host town of the facility to make sure that they know about the plan and something like this with a major amendment that the Casela will be, they'll have to make those parties aware. So we just double check with them. Act 250 will likely get involved, but that's generally not something that we heavily get involved in. I see that Mike Casela is on the phone. I'm wondering if he can is interested in responding to that question regarding act 250. I'll just give Mike a second here if he doesn't want to. Yeah, we're working through it with the state and all the permits that we need and are required. Thanks. And generally it's a conversation between us and the facility owners too, just they can answer any questions or concerns that we have on the permit side of things. But given that they did receive approval for managing an increased amount of organics through the facility at the end of last year, I don't anticipate there being many issues for them on these changes that they're proposing on this round. Great, thank you. Other questions? All right, Sarah, I see there's a motion. There is, yeah. The motion is the Chinden Solid Waste Districts Board of Commissioners hereby resolves that the all-cycle transfer station on Avenue B in Williston, Vermont be included in the district's solid waste implementation plan to incorporate their amended operations and facility food waste capacity as detailed in the solid waste certification application dated November 24, 2020, contingent upon all-cycle waste ink obtaining and maintaining compliance with all necessary local, state and federal permits. Second. I'll make that motion if nobody else wants to. I'll second that for Richmond. Any discussion? All right, all in favor? Aye. Aye. Any opposed? Any abstain? Great, motion passes. Thank you all. And I will send out the red line strike. I anticipate it to be up on the ENB in the next couple of days. So that'll become the official notice to go ahead. So I will share that for sure. Great. And that's the environmental notice bulletin, correct? Yeah. Thank you. Thank you. Sarah, we have the loader bid results next. Yes, we do. Let's back to you, Josh Tyler. Back on. Bring it up. So we have two loaders that were up for purchasing this year that were part of our capital budget that was approved. One for the MRF, and that's contractually required that we replace it this fiscal year based on a prior agreed upon contract. And one of our loaders at the compost facility, which was also kind of brought to the board in February saying, you know, we had a significant rolling stock shift with our new operations as we developed that. We went out to bid for two large loaders, full size, I guess I can call them loaders. When we got our bids back, each loader was specific to each use. So compost had guarding in different places. While, you know, the MRF loader had guarding in different places. The compost loader has been presented in this memorandum. The MRF loader we actually pulled from requesting a purchase because there was concern that the larger loader frame wouldn't fit into a couple of our key bunkers. It was brought up by two of the vendors that were bidding. So we looked at it because the MRF has specific tires which are wider than your standard or normal, you know, kind of loader tires that are typically outdoors. And because of that, we decided to take another look at the size of the loader and we're gonna re-bid that and bring that back to you in January. Also, the larger size loader was above what we had budgeted in our capital. We're making a jump to a bigger loader at the MRF just because they could use more horsepower because recyclings, I don't think it's getting any heavier. It's just we have a pretty full tip floor at all times and pushing and managing that is important. We will get a larger one, just not make that full jump to the largest size we could get. So we'll be back to you in January. But for the MRF loader or for the Organics loader, we have presented the bid results. We have our grading system and we narrowed in on NordTracks which is a John Deere loader. And the key components were, they can get us the loader in 60 days. Their quality of service has been very strong for us. The Beauregard equipment, which is a case and the CRW Woods equipment, which is a Volvo, that was the kind of the determining factor in our grading system that put them below wanting to purchase the John Deere. And we've had really poor experience with the Volvo dealer in their service department. The machines are good, the service is not. And we've had an excavator in and out and a mini loader that we bought through them in and out. And it has been a nightmare. It has not been, the equipment's come back not fixed. It sat in there for long periods of time without any answers. So that's why we kind of decided not to go with the Volvo and same with the case Beauregard equipment doesn't carry a lot of equipment on the shelf. Its facility is a farther away than the John Deere facility. And so because this bid came in under our budgeted amount, we wanted to stick with the John Deere. So that's the distinction there. And then the cat came in higher than we budgeted. So that's our logic behind this. Another thing we're doing with this is that we've kind of mixed and matched different loaders through the years at our compost facility. And one of the more frustrating things is we use different buckets for each machine. But there are buckets specific to each machine. So we can't interchange our larger or smaller buckets for whatever it is we're doing at this point. So a portion of this, because we budgeted $270,000, $35,000 was to purchase buckets that would be interchangeable between the two pieces of equipment. So that'll save us operational time and money for that matter that we don't have to stop one thing with one loader and drive it somewhere else to use a bucket for that specific purpose. So that was important to us to do. That's why this is a little more expensive than we typically have in for a loader. That's our logic. Did you go with a quick attach system on each loader? Yeah, it's a JRB, I think 418, which is the one we have. We just bought a loader 644 in 2017, 2018, I think. So both those attachments will be the same. So we can move between a seven yard bucket, four and a quarter, and I think a three yarder, which just basically depends on where you're at in the yard and if you need to fill the screener, if you need to fill a truck or if you need to operate on our working pad. Do you have forks, too, that going on, or? Yeah, I'm with forks. Is there much difference in how each piece of equipment, the controls the same? How much different is their training difference? They're relatively the same. So when we purchased the loader I was talking about in 2018, we actually requested anybody who wanted to be taken seriously in the bid. They brought a machine on our site and we tested it there. So we actually demoed the cat, the case, and the Volvo in 2018 and they were all decent. The operators like some over others, but there wasn't a significant difference in all of them. They all kind of have the same format. Nothing runs like a deer. You get commission on here, Ellis? No, I'm a Volvo guy. That's why I was surprised when he was saying all that. We have both down at, we just bought a brand new case and we have a two-year-old Volvo. And full disclosure, I'm pretty sure the Murph loader is gonna be probably a case just because as you can see, the case is the least expensive piece of equipment and for the Murph loader, I think I might, I'm gonna go with somebody who we haven't truly tried and tested yet. But I have heard things with case as far as their dealer support and maintenance and availability of parts that I took that into consideration as well. Yep. Yeah, and we'll do the same thing when those... They run those loaders hard out there at the Murph. I've been there to dump recycling trucks and you want a machine that's gonna last your seven-year rotation. And I agree with you, recycling is getting, it's not getting heavier, there's just more of it. I was doing some calculations and our monthly totals have gone up 40 tons since COVID started. But he's home and recycling at home. Per month. Yeah. Oh, that's impressive. Oh. Yeah, these are pretty heavy production uses so we can't really afford downtime for either of them. That's an important thing for us. Are there any questions from commissioners? Josh, I have a couple of questions. Is there a warranty included in the six expense summary here? Yeah, it's a standard warranty. I couldn't rattle off the numbers off of the bat. We always ask for an extended warranty and see what it costs and what it looks like. It didn't seem like it was worth getting the extended warranty. And we also have our internal maintenance program. We'll keep up on a lot of the stuff as well. And was there any sense for, I don't know, I'm just thinking about estimated operating expenses, fuel maintenance, et cetera. They're all about the same category. They're all tier four EPA standard diesel engine. One of the things that I am excited about, which won't happen now is that a lot of these guys are looking at maybe a hybrid option. And I think the hybrid options are coming out the next year or two. So hopefully in our next round, our next cycle, I would love, especially at our compost facility, because we've got such a large roof area to have a charging station run off a solar and to charge a hybrid loader. I think that would be fantastic. But talking with multiple vendors, we're not quite there yet. But as soon as somebody cracks that nut, that's the next thing up. All right. I don't have any other questions. Sir, unless there's any further discussion, sir, do you want to read the motion? Yep. Here is all that the board of commissioners authorizes the executive director to enter into a contractual agreement for the purchase of John Deere 644L front end loader from North Tracks Equipment from Williston, Vermont. And that should be four for an amount not to exceed $255,000. So moved. Second. Any discussion? I think we got a lot of questions in there. All in favor? Aye. Aye. Any opposed? Any abstain? Motion passes. Thank you, Josh. Thank you very much. Thank you. All right. So at this time, we've gotten through the bulk of the agenda. We're now ready to move into the executive session. Sir, are any of you be willing to read the language that would be wonderful? Yeah, if you have that. I do. I move that the board of commissioners of the Chittenden Solid Waste District go into executive session to discuss ongoing and pending litigation where premature general public knowledge would clearly place the district, its member municipalities and other public bodies or persons involved at a substantial disadvantage and to permit staff in the Solid Waste District attorneys to be present for this session. Right here, motion. So moved. I'll move that. All right. Second. Thank you. All in favor? Aye. Aye. Any opposed? Any abstain? All right, we'll see you in executive session. You and not the John's. Thank you. All right. Just let me know when we got everybody. We definitely have a quorum. All right. I entertain a motion to leave executive session. So moved. Second. All in favor? Thank you. All in favor? Aye. Aye. Any opposed? Any abstain? All right. We now have left the executive session. Any other items for discussion this evening? Hearing none. I just want to, before we go through the motions to adjourn, just say thank you all for sticking with us these last eight months through the virtual meetings and adapting to multiple platforms and sticking in there and just hope you all have a safe and happy holidays and we'll see you in the new year. So at this point, entertain a motion to adjourn. Last six months. I've gotten a couple hands. Logan? No, Tim, I'm just, Tim, Tim, Tim. All right. Just a question, a follow up question on Sarah's memo about schedule. Did that require any action? No action needed. It was just a notification. Thank you. But thanks, Abby. Question. I just wanted to thank Sarah for that long explanation and I really appreciated her respect for the staff of CSWB. Yes, cheers. Thank you, Abby. All right. Thank you, everyone. Thank you to the staff. Thank you, Sarah. I have a second on that. I just not heard Alan as a first. Second. Second. Well. Second. Do we need to acknowledge that there was no action taken as a result of executive session? We do not. Okay. But great question. So we have a motion. We've got a second. All in favor? Aye. Aye. All right. Happy holidays. And it's Sam. All right. Happy holidays. Thank you. Take care, everybody.


So I'm going to give five lectures, and somehow the connecting theme through all of them is going to be descendant integrals, virus hour constraints. And so I thought I'd start with the simplest place, which is for curves. So if you said last week you had surfaces in three folds, we start here with curves. And the very first part is with the cotangent lines. This is the beginning of many of the themes and the subject. And I thought I would start inside five lectures to start with a relatively careful discussion. So the first topic will be about deformations. And oh, I should say that I had sent Andre the link to these slides. So if you want to have access to them, he put them somewhere. And then we'll drop your link in the chat. So we start with curves and we're going to consider curves. And they're allowed to have nodes, so they're nodal curves.  And for us, the curves will be complex algebraic varieties of dimension one. And when I say nodal, I always mean at worst nodal. That means that a nodal curve may be non-singular. And the node looks like this. So this should be relatively familiar to everyone. So these are our nodal curves. And we will assume some aspects, some kind of notational aspects. Our nodal curves will assume to be connected and complete or projective. But they will not be assumed to be irreducible. They will typically allow it to be reducible. So a picture might be something like this. It's a possibly reducible, connected, complete nodal curve. And we will start with some basics about the deformation theory. And it's not my goal here to give some exhaustive treatment of the deformation theory for curves or nodal curves, which is not incredibly hard, but still would take some time. My goal here is to show you the appearance of the cotangent lines that come in that immediately. It's that you can't resist them. So we start with C, a nodal curve. And then here's our C. And it's a nodal curve, as we said, following the conventions I just explained. And if C is non-singular of genus G, this is the easy case, the best-behaved case, then the deformation theory, the deformations, infinitesimal deformations of C are given to H1 of the tangent bundle. And you can calculate how many dimensions H1 of the tangent bundle has using Riemann-Roch. And you get this 3G minus 3, which is the dimension of the moduli. It's a calculation that goes back to Riemann. And what's very important here is that the higher the H2 is 0, so that these are all unobstructed deformations. So this says this moduli space is smooth with a dimension 3G minus 3. So that's the case that, well, it's a smooth case. And there's no tangent lines there. So they come up in the discussion of the nodal curves. So if C has nodes, the deformation theory is slightly more complicated to describe. You have to, well, it's an axed. Actually, the deformation theory is natively always an axed. It's kind of an axed, then it's an H1 here. Because this scalar differentials is locally free in the non-singular case. But in general, this sheaf of scalar differentials is not locally free. And the actual deformations are given by an x between the scalar differentials and OC. So that's the deformation space. And again, this x2 is 0. You can check this yourself. And the deformations are, again, unobstructed. So the deformations of nodal curves are also unobstructed. But still no cotangent lines or tangent lines. So where are they come from? So the deformation theory is going to be more interesting for nodal curve. Because somehow, if you have a nodal curve, so here's your curve, it's a nodal curve, the deformations have deformations away from the node. And they also have deformations at the node. The deformations at the node can either preserve the node or smooth the node. And that's where we see the tangent lines appearing. So how to say that precisely is here. So nodal curves are more interesting. And here I label, I name the nodes N1, 10 delta, the nodes. And as I've said before, that the full deformations of this nodal curve are given by this global x. But there is a local to global x. There is a local to global x sequence. And I've written it here. And on the right-hand side, so the local to global x sequence expresses some global x, which is what we're interested in, in terms of some local x's. And this global x surgex onto H0 of this sheaf-axed 1. This global x is the deformation space. And as I said, this local to global x sequence has a certain quotient here, which is H0 of this sheaf-axed. And the sheaf-axed, since the Kepler differentials are locally free away from the nodes, is actually a skyscraper sheaf supported on the nodes. That's what that sheaf-axed is. And the main calculation you have to do here is you have to calculate what this sheaf-axed is. And as I said, it's 0 away from the node, so it's a local calculation at the nodes. And one thing that's easy to see is just by basic commutative algebra to calculate this sheaf-axed, as you can easily see that this is dimension 1 at every node, that it's not only a skyscraper sheaf supported the node, but it's actually at the node, it's just one copy of C. So that's easy to do. And then if you do that calculation with some concentration, so that the question is, what is that copy of C? Like, which C is it? How to identify that copy of C intrinsically in terms of geometry of the curve? And the calculation that's more interesting to do, and as I said, if you want to understand it, you one has to really do this oneself, is that this x at C is canonically that the one-dimensional space, which is the tensor product of the tangent lines in the two branches of the curve. If I have this nodal curve at the node, there are two complex one-dimensional spaces, which are the tangent spaces at the two branches. And this x is the tensor product of these two tangent lines. So this is where the tangent lines appear, so somehow immediately in the deformation theory. So I encourage you to do this local calculation for this x and to identify with the tangent lines. And so we have this sequence from the deformation theory of C. That's the x to the sum for every node. There's a map to the tensor product of the tangent lines. And moreover, it's surjective because of this local to global x sequence is surjective there in this case. And what is the meaning of this sequence is that this x1, this sheaf x1, is about the deformations in the nodes. And if the global x maps to a 0 in the sheaf x1, it means it's a deformation that preserves the nodes. So one way to interpret this is that the kernel of this map is the deformations which preserve the nodes. So this is really the first occurrence of these tangent lines. And we will see them several times, but it's good to understand this first one. I thought that since it's a course, we could start at the beginning. So that's basically the foundational discussion here. And it leaves you, if you want to really understand it, you have to do this crucial local calculation and identify the result with the tensor product of the two tangent spaces, which I encourage you to do. OK, so then we go to Moduli, that if I look at the Moduli of curves, that's this non-compact mg, then this is the moduli space of non-sing of the curves we've already seen. This has dimension 3g minus 3. And more relevant for us for most of what we are going to discuss is actually the compactification by Deline Mumford, which is mg bar. And this allows nodal curves. They have to be connected, as I said, and the canonical bundle should be ample. That's the stability, the Deline Mumford stability. And we can also have points. We can take the moduli space of genus G, and pointed non-singular curves. And also there's the compactification mg and bar. And here we consider that when we have some positivity there. And an element here is the curve is nodal and connected. And we have N mark points. And these are distinct points, the non-singular locus. And the stability is written here as the canonical bundle twisted by these points is ample. OK, so I hope many people have seen some aspects of the moduli space of curves before. The point is the compactification is non-singular reducible. I guess I should also say compact, I already did it there. It's a proper Deline Mumford stack, or it's an orbifold if you like that language better. And the dimension is 3g minus 3 plus N. And the rigorous treatment of all of this theory of moduli happens in mathematics in the 60s through a combination of different things. But you can say Deline Mumford in the 1960s. But the roots of the ideas go back to the 1860s, the Riemann. All right, so we've seen the tangent lines. And I wanted to show a little bit how they come up geometrically now. I showed how they appear in the deformation theory. So for every mark point, that's this I. That's the index for one of these N. So this I is one of the N. For every I, there is a line bundle, a complex line bundle determined by the tangent space at the ith point. So if this is the curve, it has an ith point. And the ith point has a tangent space, T, C, I. And if I put all those tangent spaces together, it gives me a complex line bundle over the moduli space of curves. And I have actually N of them, one for every mark point. And these mark points are never allowed to be the node. So there's never any confusion about what the tangent line is. It's well-defined. So these are line bundles on the moduli space of curves that come from the intrinsic geometry of the moduli space. And one way they make their appearance, I already said that they made their appearance of the deformation theory. Another way to see that more geometrically, perhaps, is if I look at the locus of stable curves, where, for example, I have two curves attached at a node. So the left hand one has genus G1, and the right one has genus G2. Then this diagram of attaching two curves at the node can be put together in moduli and giving a map from the two generon, the halves of the curve. You take the Deline-Mumford spaces, and you attach them at the points and make a bigger genus curve. And if you count dimensions, the dimension here will be one more than the sum of these two dimensions. So one thinks of this map as somehow being a divisor. And it's almost always a divisor, but sometimes it's not injective. But anyway, this map is co-dimension one map. And you can ask, since all these spaces are non-singular, you can ask, what is the normal bundle of this map? And if you understood what I said about the deformation theory, then you'll know what the normal bundle of this map is. The normal bundle of this map is exactly those directions which correspond to the smoothing of this node, because all the others are already absorbed here on the left-hand side. So the normal bundle comes from exactly the deformations which smooth that node, and we have explained what those are. They're the tensor product of the two tangent lines. So these tangent line bundles are very useful in the geometry because they tell you, for example, what the normal bundles of these divisors are. Let's start. Okay. So those are the tangent lines. And of course, if I take their dual, I get the co-tangent lines. So actually I'm getting a little bit of an echo. So maybe one could try to silence the microphone there. Okay, so if I take the dual of the tangent lines, I get the co-tangent lines. That makes sense. And another way to say it is these are the, this is the complex line bundle that's determined by the co-tangent space at the i-th point. And we'll use the more traditional notation instead of using this TI star, we'll just use LI. That's more standard notation, the subject, the co-tangent lines. And I would say that in geometry, very often co-tangent lines are used instead of the tangent lines. And one of the reasons is that this is the first result is that this line bundle LI, the co-tangent line bundles NEF on the Deline-Mumford modular space. So this is the first geometric result about the co-tangent line. And I thought I'd give some kind of proof of it. So the co-tangent lines have some positivity. That's this NEF-ness. So if you don't remember what NEF is, it means that if I intersect it with any curve, I get something that's greater than or equal to zero. Okay, so I wanna give a proof of this NEF-ness. And so let's start with a curve. So we're gonna take a curve, a test curve here. I take a one-dimensional base B that's here. And I want to consider this B mapping to the modular space of stable curves. And how do I arrange such a map? Well, this is more or less the same as giving a family of stable curves over this base B. And I have to then also give the sections, N sections, which give me the points. So if I have this family of nodal curves over B with these N sections, such that every fiber is a stable Deline-Mumford stable curve, this gives me a map from B to MGN bar. So if I wanna test the NEF-ness, I have to prove some result about every one of these families. So let's start with one. So then of course we have to use something. And the positivity we use is that for this moduli, we can look at this space. The space is omega to the, it's the, yeah, so maybe I should put the K another place here. It's the dualizing sheaf twisted by these sections and I take the Kth power of it. This gives me some for high K, this gives me some vector bundle on the base, this R zero pi lower star. And it's one of the basic results is that this bundle is semi-positive. And this has been explained, this is explained very nicely in this paper by a Yanush Kolar called projectivity of complete moduli. That's a, that paper has a goal of trying to construct these moduli spaces without using GIT. You have to get positivity from somewhere without using GIT. And he gives you a proof of this semi-positivity. So I recommend looking at it in section 4.7. So we have to start with something. So this is some vector bundle at semi-positive, which means that at least it's of degree greater than or equal to zero. So if you look at growth and decrement rock applied to this, you'll find that that implies that the first churn class of this omega pi twisted by the sections, the square of that is greater than or equal to zero. The semi-positivity for this K implies this result. That's by growth and decrement rock applied to that surface. So this gives us something. This is our start, but it doesn't look like this has anything to do with the cotangent or tangent bundles, but it does by a kind of a clever idea. So we take, this whole proof is about, so here's this curly C, which is actually a surface. It's a surface because it's a family of curves over a one-dimensional base that gives us a surface. So this whole proof is going to be about some intersection theory of this surface. So the first thing we've learned is this first churn class of this particular line bundle, that square is greater than equal to zero. Now, the second geometric state, but if I take that same churn class and I multiply it by the class of the section, the sections give me some divisors in the surface, S1, S2, that's the locations of the mark points. If I take the intersection of this first churn class of the bundle we've decided whose square is positive, if we take that intersection with a section, the answer is that this is exactly zero and that's for geometric reasons. That when I take this, it's the degree of this dualizing sheaf on the section and then also the degree of the section restricted to itself. There's no intersection of different sections for I not equal to J because the sections are distinct. That's one of the assumptions. So if I look at what this is, this is the cotangent line at the section and this is the tangent line at the section. So we get geometrically that this intersection product is zero. All right, so what we have here is we have this self intersection is greater than or equal to zero and the intersection of that line bundle with the section is exactly equal to zero. So now we wanna use the Hodge index theorem and one has to concentrate a little bit when one uses it here. So the Hodge index theorem says on the surface if I look at the narrants of area group that I get the signature that's minus, that's plus one and then all minuses. That's the statement of the Hodge index theorem. And so if we go back up here, we can assume suppose, oh, sorry, I'm not sure what happened. Suppose this was positive. So this is greater than or equal to zero. So suppose it was positive, then it would eat up the plus in the Hodge index theorem and moreover, the section is orthogonal to that positive element. So it would have to be in the minus part. So if we had a positive there, then the section would have to be the sum of the square of the section would have to be minus. And then if you go understand what the cotangent line is, it's actually equal to negative because this is the tangent line. So this being minus means that this is plus. So that proves that the cotangent line intersects this curve positively. Of course, but that may be not the case and maybe this we have a zero here. It could be that we have a zero. So we handled the plus here, the plus here gave us a minus there which gave us a plus here, but it could be that we have a zero here. Then we have to see what happens. Well, could we have a plus here? Have a positive, strictly positive. That would contradict the zero here because if this were strictly positive, then this class would have to be in the minus in the Hodge index, but it's zero. So the outcome of that outcome of this is that if you go through that logic carefully that the fact that this is greater than equal to zero and intersection with SI is zero means that this SI squared is less than equal to zero and the intersection of the cotangent line is minus the SI squared so that's greater than equal to zero. So one of the puzzles in this proof is could zero actually occur? And the answer is yes, it could occur. And that's a kind of question, look at reducible curves. Okay, so I hope that was understandable. Maybe one has to go through this logic oneself, but the steps are to use this semi positivity of our pie lower star and then use the logic of the Hodge index theorem. And there's another wrinkle in this proof which is that you might think, well, am I really using this? Am I legal to use the Hodge index theorem for these surfaces which have many components and all that? And so you have to think about that yourself. You can take them apart and apply it to one component at a time if you have to. Okay, so that's the first geometric result. These cotangent lines, they're actually neff which is kind of a nice thing. So it gives you some positive line bundles or weekly positive line bundles in the modular space of curves. And the standard notation for the first trend class is this psi, psi i is the first trend class of Li. Can I ask a quick question? Yeah. Could you repeat where the tangent and cotangent lines come into that proof? Yeah, okay, that was a little fast. So maybe I'll make a little, some kind of new page that might help if I know how to do that. So somehow in the full screen, okay, I have to understand how to do that. Okay, I can still, I can explain it here. So we have this family of curves. That's the curly C over the base B. And we have these sections. Okay, so let's look at section one. So the first thing, one of the things that's coming here is the self-intersection of that section. That was one of the actors in this argument here. And if you ask, what is the self-intersection of a section? Geometrically, that's given by the churn class of the normal bundle of that section. And what's the normal bundle of the section? Normal in the section is the tangent space, the vertical tangent space. And what is that? When I take the normal bundle of the section and pull it back to B, that's the tangent line. So in some sense, you can say here that this intersection here is the intersection of the base with the first churn class of the tangent line at R. So it's the negative of the cotangent line. So that's one way where it comes. If you look at the section and you look at the self-intersection of the section that has to do with the normal bundle, the normal bundle has to do with the tangent line. Another way that's coming in this proof is that if I look at this dualizing sheaf and I restrict it to the section, the dualizing sheaf to the section is everywhere the cotangent line. So we can say this is like the cotangent line and this is like the tangent line. Okay, I hope that helped. So there's something slightly confusing here always about modularized that in modularized space we consider structures on the modularized space but those structures give actually geometric structures on every family. And part of understanding this is how to translate the abstract structures in the modularized space to the specific geometric structures in every family. All right, I hope that went some ways to answering that question. Rahul, we have a question in the Q&A. What kind of singularity can seed have? I mean, the curly C? I think the curly C, yeah. Yeah, so the curly C, I mean, every fiber has a nodal singularity. And the kind of singularity C can have is something like X, Y minus T to the K, things like that. Because you can have some deformation of the node. You could have a smoothing of the node. Another way C can be singular is it could be, it could just be like this where it could just be a nodal curve and then C could have two components. So that's also possible. That's also a different kind of singularity. So it can have this kind of smoothing of a node singularity or it could have just, it could just preserve the components. The technical definition of, I mean, we can take here B to be non-singular and the technical definition of C is it's a flat family of nodal curves over the base. So it can have any singularities are allowed by a flat family of nodal curves over one dimensional base and they're kind of two kinds. There's this kind and then there's the one where the node is just preserved. So yeah, I mean, I didn't give every single detail in this argument. So if you look at the Hodge index here and that's normally stated for some non-singular surface. So you have to make some, you have to fiddle with that a little bit to put it in that. This is some technical moves which are kind of standard. All right, so the standard notation here is that this psi I is the first trend class of the cotangent line. This is completely standard in the subject. The psi I is the trend class of the cotangent line. And we have a forgetful map from MgN plus one to MgN or you forget the last marking and then you can define the Kappa classes. The Kappa classes are given by taking the homological push forward of powers of the cotangent line. So these give us some classes in the modular space of curves. And another basic geometric fact is if I take the Picard group with the modular space of smooth curves this is geometrically generated by this Kappa class and the cotangent line. So again, you can't avoid these cotangent lines. I've given you many different ways to think about them and you can't avoid them. They're just basic aspects of the geometry. So nefness, that geometric nefness with the argument I gave you using the Hodge index theorem is that by this geometric nefness, if I take the churn classes of the cotangent lines and raise them to powers that the answer is always greater than or equal to zero. And that's because if you take nef line bundles and you take their powers you always get something greater or equal to zero. That's an aspect of nefness. And of course the integral vanishes unless we have the dimension constrained. We have to have the dimension of the integrand match the dimension of the modular space which I remind you is 3G minus three plus N. So if you're interested in these integrals of the cotangent lines, which we will be then the one thing we know is they have to be greater than or equal to zero. And as I said, of course they vanish if you're not if you don't match the dimensions but a question that emerges directly from the geometry is can these vanish if the dimension is correct? Like can we actually have an actual honest intersection of cotangent lines, which is actually equal to zero? That's not ruled out by this Hodge index argument. And actually I don't know how to address this question strictly speaking geometrically. We'll give an answer to this, hopefully at the end of the lecture and the answer is going to be yes, it has to be strictly positive but this nefness doesn't quite imply it. And as I said that it's not if you want one wants to understand this it's good to find loci where this cotangent line is trivial and you can find that. Like an example is if I look at Mg1 and if I look at the locus, okay maybe let's do it Mg2. If I look at the locus where the two points bubble off so this is a genus G curve and the two points are on a M03 bubble, then this gives me a big it's a divisor in fact here. It's kind of one of the divisors I explained before but it gives a divisor in Mg2. And if I look at how the cotangent line at one of these mark points behaves, it's just trivial because this cotangent line is the cotangent line of this mark point on this M03 so nothing is moving. This is a pretty large locus where the cotangent line is trivial which is okay for somebody who's nef. But one, as I said, this leaves this question is that can we take away this or not? So far I've been giving a lecture about kind of elementary aspects, the geometry and it's true that I skip many steps and it's good to think about all of the things I'm saying. So it's not a bad idea if you want to understand the subject to one to do that X calculation I explained at the beginning completely honestly so you believe the canonical correspondence with the tangent lines. The second is to go through this Hodge index theorem argument and do it completely honestly in terms of the components and singularities. And then after that you'll come to this point where you think about these cotangent lines and there's various way classical ways to try to compute these intersections but then now comes a leap in the subject where I'm gonna tell you how to compute these intersections. And this is something that's very non-trivial and it was some developments that happened now quite a long time ago. So this is when more or less I was in graduate school. So it's in the early nineties and that's Witten's conjecture and Witten tells you that his conjecture how to compute this, how to compute these intersections and to explain the form of the conjecture that he wrote I will switch to his notations which is this notations of the descendant insertions that's these towels and the bracket and that's very simple notation so if I write this bracket and this towel insertions then this is just equal to this intersection of these cotangent lines. So when I see a K1 exponent that goes as a towel K so somehow towel J corresponds to some point to the J. This notation is kind of nice because it captures the symmetries of the integrals here is the basic symmetry, the SN symmetries. You don't have to really write all of these subscripts. So they've been taken out here and this genus here tells you what genus is. This N tells you how many points but it's kind of redundant because you can just look at how many points there are here. So that's often dropped and the reason he writes it like this is because this is viewed as some insertions and some quantum theory. So as I said that this thing is zero if the dimension constraint is not satisfied and it's also zero if we have some or not unstable curves but for the most part as long as you're not violating dimension constraint then this is geometrically well-defined. All right, so the state witness conjecture we form a generating series. So that's we fixed it. Well, to start we fixed the genus and the generating series has some variables and there's one variable for every non-negative integer that's this zero, one, two and these variables will course, I mean, there are index, these variables index these tau insertions. So there's a one variable so to speak for every tau insertion and that's kind of exponential generating series. So it's a definition as you sum over all sequences of non-negative integers with only finally many non-zero terms. For every one of these sequences you get some bracket here and you put the generating series, the variables in this way which is kind of exponential way. And as I remind you that this bracket is this integral that we're interested in. You can write this more efficiently or you can just write this just on the face of it is exponential where you view this field fives formally all of the towels with the variables TI. So that's a more efficient way to write it, but nevertheless that's a, you can convince yourself that this series has coefficients which exhaust all of the integrals all of the descendant integrals that we're interested in all of these integrals. Where did they go? Here they have all of these integrals appear and nobody else appears. So in some sense this generating series exactly generating series of these cotangent line integrals. Okay, so that's so far nothing's happened. I've just written down the series which carries these integrals and we can put all of the data for all the genera together. And then we need an extra variable for that. That's this lambda. So now we have this T so this T stands for this sequence of variables T zero, T one, T two. And if we put all the genera together we need another variable which keeps track of the genus and that's this lambda. So we have F lambda T and for the genus zero one has to be just a little careful because there were some that don't make it from our point of view that don't make any sense. We don't consider a genus zero with one and two points since these are unstable configurations. So those integrals if they appear here are just set to zero. This was already said somehow in these it was said somehow here that also zero if two G minus two plus N is less than or equal to zero. Okay, so we can put all of this data together and we get this function F and then there's the notation for partial derivatives which is kind of very natural in this notation which is that if I put double brackets that means I take partial derivatives with respect to the variables associated to those tiles. Okay, so then Witten's conjecture which was proven also in the 90s by Kinsevich. So soon afterwards actually is this series of differential equations as written as KDB form. And it says that for every N greater than or equal to one we get an equation and this equation well it's about these derivatives of F. So this is some non-trivial equation and whatever it is it must be saying something about the cotangent line integrals because F is just made of its coefficients or just the cotangent line integrals. But so Witten gives you an infinite sequence of integers and this infinite sequence of integers is related to this KDB hierarchy and I want to explain at least one origins of that. And it's just extremely explicit thing here. I think I've explained all aspects of the notation. So if we look at the first one it's standard here to let you be the second derivative of F with respect to T naught.  And this is exactly this KDB equation and it's a mathematical origins. That's the very first one that's the N equals one equation here. This is the KDB equation as mathematical origin comes from the study of shallow water waves. And this appears in the 19th century to describe some persistent waves on channels and in the Netherlands. And here I tried to give you the dictionary that this tau zero is the spatial coordinate, T1 is a time coordinate and U here is the height of the wave. And it's a very natural question to ask like what are waves? What is this? How is it that we have these shallow water waves coming onto the modular space of curves? That's kind of an amazing thing. Okay, so I'm gonna say a few words about that but not many, but so I would just like to say what I've done here what actually Witten has done with Witten conjectures is that we take these cotangent line integrals we associate a generating function. This is this free energy and then it happens that they satisfy this amazing sequence the infinite sequence of partial differential equations that the first one is the classical KDV equation and the higher ones are this KDV hierarchy. And so it's a natural question to ask what do water waves have to do with MGM bar? And this is a long and interesting story and I recommend that if you want to read about it this is paper of Witten, this is around 1990 maybe 89 or something like that but well into the past century which is 2D quantum gravity and intersection theory and modular space. And the roughly this kind of a high level outline of what happens there is that there are so there's a theory of integral systems and matrix integrals the integrals over spaces of Hermitian matrices and what Witten the connection Witten makes in that paper is that one way to think about these matrix integrals is mathematically analyze them and it was known before Witten that such matrix integrals they're generating series give rise to some functions that satisfy the KDV equation that's a theory of integral systems but the really new connections made there is that Witten gives some geometric way of thinking about the same the matrix integrals in terms of what he calls 2D gravity which is some integral over all Riemann surfaces and when he tries to identify what the specific integrals are what they should correspond to in the 2D gravity he gets exactly these cotangent lines. So if you believe these leaps here that this theory of integrals over the cotangent lines can be written explicitly in terms of some matrix integrals then there's a second leap those matrix integrals are connected to KDV equations and this was proven just shortly after by Kinsavich and he does exactly write these cotangent lines in terms of matrix integrals slightly different matrix integrals that's called Kinsavich's matrix model but essentially goes by this path but that's an amazing story and if you want to look at the first pass through it I recommend reading this paper of Witten's or these parts of it. So that was the very first pass the second pass goes through Hurwitz numbers and there's different ways to go through Hurwitz numbers and I'm not going to explain that here so much but the results can also be proven using Hurwitz numbers instead of this matrix integrals and KDV equations you use relate the cotangent lines to Hurwitz numbers and then use some properties of the Hurwitz numbers to show that they satisfy these equations actually you can use them to prove the various matrix models also exist by the Hurwitz numbers methods and then there's the actually I think that in some sense algebraically simplest approach if you want to like the opposite of the beginning if you want to look at the simplest path I recommend this paper by Kazarian Lando in 2006 but anyway those are three things if you want to learn a little bit more about this you can read the Witten's paper or you can read this paper I wrote with Andrei Okonkovan from Witten 3 Hurwitz numbers matrix model but you have to like Hurwitz numbers for that and the most efficient proof also using the kind of Hurwitz number path we take is by Kazarian Lando 2006. Okay, that gives us some clues and the question is can we calculate? So that's got this infinite sequence of equations can we actually use them and I wanted to give, and the answer is yes and here I wanted to give a very first example which is this tau one in genus one that's integral of psi one over M11. So M11 is the space of elliptic curves with one point and this space is one dimensional and normally it's given by some region like this quotient by SL2 but here we take the compactification so it's one dimensional algebraic variety or Deline-Monford stack and then it has one point and I can take the cotangent line and so in some sense this is the first high and there's the whole genus zero story but this is the first kind of non-genus zero integral think about kind of interesting one and you can ask does how to calculate the value of this using this set of equations and if when you first tried it doesn't seem to work so well so I wanna just show what happens. So first you have to decide which equation to pick so I tell you pick n equals three and then if you write this n equals three equation and you set lambda equal to one and all the ti is equal to zero you set these after the differentiation of course then you get this equation. So that has some genus one on this side some genus one on that side some genus zero, genus zero and it has lots of tau zeros and it doesn't look like it's actually helping us too much but there is an equation that then helps us it's called the string equation the string equation tells you how to remove tau zero it's a very easy equation and it's on the next slide but what it does is it tells you how when you have a bracket how to remove a tau zero and it says more or less I take this tau zero I remove it and I bump this down and I get the tau one and I said the string equation I wrote it on the next slide but the string equation tells you how to remove this tau zero and it's simple let's you simplify this equation that's come straight from the KDV equations and if you apply those if you apply the string equation to simplification then you find this very nice equation and this very nice equation you can solve and you get this rather well-known result which is the first churn class of the contangent line in M11 has a degree one over 24 and so this already shows that in the subject the orbital nature plays a kind of decisive role there's fractions everywhere and the reason for this fraction is because the modular space M11 is not the right way to view that modular space is an orbital not as a space and that's these automorphisms that are giving you this one over 24 so the string equation what is that? So here's the rule for the string equation it says that if I have a tau zero and I want to remove it from the bracket I just remove it and then I go through everybody else in the bracket and I lower them by one I sum over the choice of who I lower by one and then I lower that person by one so that's the rule and you can look at that rule and I've applied it several times years you can go look so here there was a tau zero and I lower there's only one person lower because I can't lower zero to negative one so I lower this three by one and by one and I lower here so you have to just play with it more interesting question is why is the string equation true? And I gave a whole proof of it here maybe I don't go through the proof now but the proof is a proof using just the geometry of the modular space of curves it's a different level of complexity meaning that the string equation geometrically is very, very simple this equation there I said there's various approaches to them none of them are very simple this equation is some kind of very deep mathematical equation the string equation is not it's an easy one so I actually gave the whole proof here but I don't want to give it now you can read it yourself or maybe you can discuss it in the problem sessions it's one of the parts of the subject if one wants to understand how Witten's equation work it's good to have a good mathematical mathematical knowledge of the string equation and I think I will skip these two slides it's just a calculation and I did it kind of carefully here and you can look at it so now some exercises use the string equation to calculate the gena zero descendant integrals that's all the descendant integrals over the modular spaces of gena zero curves they're easy and are given by this multinomial coefficient and to prove that all you need to know is the string equation and one some initial condition that's easy and while we're talking about that there's another rule which is how to remove the tau one so in this bracket there are two things that are easy geometrically it's to remove if you see a tau zero you can remove it or you can take a tau one you can remove it and by remove it I mean you can express it as a sequence of as maybe a sum like in this case a sum or just one descendant integral where that insertion has been removed it's a geometric rule and these are the two standard ones are the string equation whose proof I've skipped and then there's the dilaton equation for tau one and I didn't even give the proof here but it's proven in the exact same way if you understand the proof of the string equation you can also prove the dilaton equation and these are very useful simple rules you know that it just tells you if you see a bracket how to remove a tau zero and a tau one and these rules are very appealing in some sense they are a bit nicer than this kind of gymnastics I had to do for the KDV where I have to look at the integral and have to digest this and I find that the particular fellow that I'm interested in appears on both sides of the equation and whenever you have something appearing on both sides of the equation you might wonder like what if this had the coefficient which canceled that and vanished that would be disappointing so this is a more intricate analysis although it always works it's not the case that the coefficient it's not the case that the equation runs away from you but it might be nice to hope for generalizations of these very simple equations the string equation and the dilaton equation that'd be an equation that directly tells you how to remove a tau k and that line of discussion is a very profitable line of discussion and it's actually kind of the one of the themes that are run through the series of lectures and that has to do with the Versailles constraints so the mathematical statement is that if you know the KDV equation and you know the string equation that determines all of the descendant integrals but an easier way to determine them conceptually is from the Versailles constraints and that's what I wanted to end this lecture with so we come back to this generating series for all descendant integrals that was in Witten's conjectures and then I can take the exponential of it and get this partition function Z so in terms of keeping track of the descendant integrals there's no damage or benefit for doing the exponential they're still there and you can get back by the logarithm but this has some meaning this F is for the free energy and the Z is for the partition function so then one can write this string and dilaton equations the ones that I said are rather direct geometric equations you can write them very nicely in terms of differential operators and I've written them here these formulas can already be found in Witten's paper in 1990 and then the statement of the string and dilaton equations are that if I take this L minus one operator and this L zero operator it annihilates this partition function this is just logically equivalent these two equations are just logically equivalent to what I call the string and dilaton equation with one more input some initial condition the initial condition is one here and this initial condition one over 24 and that initial condition makes its way to this one sixteenth so anyway what I'm saying is that if you geometrically derive the string and dilaton equation in the form I said then that and you know these two initial conditions then that's logically equivalent to these two differential operators annihilating Z and if you have these differential operators it makes sense to calculate their bracket and it turns out their bracket because it gives you back L minus one you know since these differential operators annihilate Z their bracket also annihilates Z so it would have been nice if it gave you something different and this can remind one if you think about these things that if I look at holomorphic differential operators in the variable U this is usually the variable Z but I already have so many Z's I switch to U if I look at holomorphic differential operators in the variable U that's given in this form curly L N is minus is negative U to the L N plus one partial partial U so these are some differential operators on the line and they satisfy this V-R-S-R bracket and if you go look what we have here is a beginning of a representation of this algebra if we can make this curly minus one goes to L one curly minus curly zero goes to L zero it's a beginning of the representation of this algebra of holomorphic differential operators in the variable U and the line which could be considered at this point just like a flight of fancy is that maybe we can extend this representation to every K but goes to what and we have some clues because there's the written and of course then we'd want to have this an annihilation property so one has some clues some road maps but the answer is the following for every positive N so that gives the L one L two L so on you define this differential operator okay and as I said that you can ask where is this formula coming from and as I said that there are some clues if you ask to find a differential operator that has well you have to constrain so to speak what terms will occur here so if you know ahead of time it's just these terms and there's some good reasons to know ahead of time it's those terms it's a good guess to make anyway and then moreover you want to impose this bracket it really constraints the coefficients and my memory is that basically uniquely constraints the coefficients up to some maybe some scale or maybe not even but anyway what I'm trying to say is that if you decide that you're going to try to find extend this representation in terms of these different differential operators you constrain the orders that you see it really constraints the possible answer I mean it's over determined systems you expect none but actually there's this guy that's kind of a beautiful thing and then you can just check algebraically like once someone has written the formula for you to check that it satisfies this VIRUSR bracket it's just some mechanical check which I invite you to do and the VIRUSR constraints then take the form that these differential operators ln annihilate this partition function for all n greater than minus one so it includes the string the dilaton and then all of these higher ones and if you look at what's going on here if you can understand what this means what it means for this operator to annihilate z this partial partial tau n it gives you a rule for removing tau n plus one just like this the string this minus one gave a rule for removing tau zero this gave a rule for removing tau one this annihilation this ln annihilating z gives a rule for removing tau n plus one and that this is zero so this goes to the other side of the equation and then if you see what happens to all these terms they're somehow lower and what happens magically in this equation is that all these terms are positive this there's one negative coefficient that's this initial coefficient and all of these other terms recursive terms are positive so what that means is that this actually proves that these descendant integrals are always strictly positive whenever so I actually don't know a geometric proof of this this is a but the this proof from this VIRUSR formulation proves that they're all strictly positive and if you want to to see what's happening here I invite you to compute this so you can say what's the proof so there's two paths to the proof the first path is this this KTV plus string equation actually implies the VIRUSR constraints that's algebraically and this was a a version of this is in a paper by Dicra for them been relented in 1991 meaning that algebraically this already implies it it's slightly non-trivial but it implies it algebraically so the metric path which is kind of beautiful is hyperbolic geometry this is Mirza Akhani so Mirza Akhani shows that if you study bounded Riemann surfaces and you look at the volumes of these Riemann surfaces with boundary then she has two ideas there one is that you can express the volumes in terms of descendant integrals and the second is that she has a recursive relation for the volumes using some decomposition in geodesics and if you connect her path it's amazing but then you get exactly the VIRUSR constraint so that's maybe the more direct way so there's two different paths okay so I'm going to stop so this corollary is for the disconnected integrals right what's the corollary I mean they are positive right no this is the connected ones I mean I'm not sure it makes much difference but when you take a look there are some signs right no but you make the connected ones out of the disconnected ones so you make the disconnected ones out of the connected ones but yeah I'm what I've written here is for connected integrals any other questions any other questions so I have a naive question what is the space of the representation power series in the T variable it's a no it's differential operators and it's like the space of differential operators and infinitely many variables because we have this that's where the LN yeah the LNs live in some differential operators on all these variables infinitely many variables but they're not just they're differential operators actually they have polynomial coefficients I mean you can constrain it a little bit but that's what we're talking about what's the structure of the representation itself abstractly is it cyclic I mean it's a I think it just gives you an yeah I mean the whole you know so I mean I think it's just an isomorphism that you know this this this algebra of whole differential operators is just isomorphic to this one but you're asking about how this acts on all of differential operators I don't know so we have some questions in the Q&A would you like me to read them to you or would you like to take them I can't see the Q&A so maybe it would be better if you read them to me another problem the first one is is it proved that the intersection numbers of psi classes belong to Q yeah they belong to Q because that's because they are so that has to do with the intersection theory they're actually integral classes on the stack and that gives you Q classes in numbers they're actually line bundles on the stack so it's churn classes will give you some classes they're integral classes on the stack and when you I mean there's z-valued classes on the stack and then when you integrate them you get some rational numbers because that's how the rules of overfolds work so the answer is yes next question is for the differential constraints is there a deformation of these differential operators that gives non-zero central charge in the differential yeah it's yeah I mean that's right this is only somehow half of the and how to put the other ones in in this context it's not so clear because we don't have some negative descendants in other contexts there's been some playing with it but I don't know how to put it in here the rest of the so we have a couple more questions about the differential constraints so one of them is would they be equivalent to what in physics is the differential of the states being physical I don't know and the last one on differential constraints is differential constraints mean that z is the highest weight vector for the vacuum module is there a similar interpretation for non vacuum highest vectors and maybe this was related to the other question we had about what the full representation is I don't know any further questions then let's thank Raul again