I'm from the University of Birmingham and I'm involved in providing the HPC facilities here. At the University of Birmingham we like a good acronym so the HPC facilities and other things we've provided are called Birmingham Environment for Academic Research. Obviously that allows us to use the word bear lots and when you see lots of our publicity and other information you'll see pictures of bears. On there in that photo are just a couple of the bears that one of my colleagues has sitting around in their office. On the other side that's our water cooled data center which opened a couple of years ago now or nearly three years ago. That's from quite early on actually when we had a bit of it you can see some of the compute nodes, some of the storage, a small amount of the Power9 AI cluster that we IBM Power9 AI cluster that we've got in. That's before all of it was there. Our group's also responsible for what we call the research data store which is providing storage across campus so that's both for people using HPC facilities but an awful lot of our researchers that use it don't use HPC at all. We've also got high speed networking setup that's for people who say either say in the medical research area doing gene sequencing. We've also got some the physics type equipment that's producing lots of data and chucking it back into the store. We've also got a GitLab setup internally for researchers who want to keep their Git repositories local. Our computer, our HPC machine, continuing the theme of bear is called blue bear. This is available to any researcher at the university. I have lots of bioinformaticians, a good chunk of engineers, computational chemists, physicists but a lot of what goes on in our system is high throughput computing rather than high performance computing. We do have a chunk of high performance computing type uses but we often see lots and lots of jobs and lots and lots of package requests from those bioinformaticians. Our system, we have the section of it that's available for free to all researchers but also research groups can come to us and ask to buy additional resources. That's either the groups that use lots of computational resources might want to have their own dedicated nodes as resources to them or they might have some extra requirements to do with particular dedicated access to GPUs or to large memory systems and where the available shared area it just isn't sufficient for their needs. This is the third generation of blue bear. This part of blue bear as we've actually been going which is a follow on from version two started in 2012 and what we do is we do rolling updates where of both the hardware on the software side of things. So we say it's the same blue bear but now actually we've got no original components from the 2012 iteration of it but in some way some of the components we've got now overlapped with some of the components we had there. We've now gone through two different storage systems that have become end of life and been decommissioned and so on. What do we have available in our system? Because of that we don't have separate clusters. So we have one login system and from that you can access a bunch of different types of CPUs, some with GPUs to make jobs out. At the moment we're running a mix of central seven or mainly central seven and we have Red Hat on the IBM power nine nodes. The sandy bridge I've crossed out there that was what we decommissioned last year. They'd become end of life so we decommissioned the last of those about Easter last year. At the moment most of our GPU our Intel based GPUs are attached to either Haswell or Broadwell CPUs or in fact all of them. So Cascade Lake we actually have two generations of Cascade Lake where the difference is in the type of interconnect. ConnectX5 versus ConnectX6 I think that's correct. Networking is always a little bit of somebody else takes care of that and tells me what we've got. We also provide a cloud infrastructure which is effectively where we allow the researchers access to equivalent of a compute node but has a virtual machine instead and these run a combination of either CentOS 7 or Ubuntu 16. They tend to have LFS and no Infinity band connections on them. What does our software stack or what sort of going to go with what we installed last year? We did over a thousand installs last year which is and we did over a thousand the year before. The image there was generated by one of my colleagues I'm just quickly reading his notes. The word cloud shows all the applications we built with the font size proportional to the number of modules were given an application name. The OSU micro benchmarks and HPL are quite big because I spent a decent chunk of last year trying to work out where debug some problems both with MPI and with our compute nodes and installed on every single tool chain we had. You can also see some of the things I rely on and TensorFlow that are some of our bigger users or bigger use cases. I've also split out there the where our installations are in terms of the easy build tool chains. You can see we're mainly using the FOS and FOS CUDA area tool chains. When I put FOS and FOS CUDA there that includes any of the things so FOS includes the GCC tool chain and the Compute tool chain those sub-chains and similarly for anything else. I've also put tool chain generation for the easy build tool chain generation for there. You'll see that well over half of our installations were against the 2019 B and again when I say 2019 B there I also mean whatever the underlying GCC call version was there I think is 8.3. We did end up doing more installations against 2020 A than I was expecting. Due to CUDA and some other issues we were looking at jumping to 2020 B and ignoring 2020 A but about September I think it was somebody asked us for our version 4 which in the easy build was in 2020 A so that I think accounts for most of those 254 by itself particularly when I think the other thing was asked for was a TensorFlow version against it. There's a TensorFlow version in that tool chain as well. So easy build at the University of Birmingham we've been going since 2016 with easy build which is about a year and a half before I joined the team in fact none of the current people building software was there when we started with easy build we've inherited it and changed it since then. All of our installations are by request so we don't preemptively install anything anymore we just wait for our users to ask for it we get quite enough installation requests in that method. In conjunction with talking with the user we decide what tool chain we're going to install it against that allows us to sort of manage where we're installing and not installing in too many different areas and for sort of our own sanity we tend to have only two active tool chain generations for installations at any one time so currently we'd only be installing in the 2020 A and 2019 B tool chains that's there. I mentioned that we had the different CPU types if you're on an Intel system and you module load something it will load a version that's being optimized for that CPU type that's transparent to the users they don't know what's going on they don't see what's going on it just happens you know we said the module passed correctly we have a few users that self build their software that's outside of easy build using the obviously modules from easy build and where they're looking at running on the different CPU types we talk through how to correctly set that up so that when they're installing they're all building their own software across different CPU types they'll end up with an optimized version for each CPU and we find that to be quite successful and the users there understand there. Talk about a few of the things we looked at doing last year um one of the major things was we were looking at doing an OS upgrade where we'd take the CentOS 7 go to 8 or write out at 7 to 8 we did all the installation work for there or nearly all the installation work we decided we were going to use tool chain 2018 B for the x86 systems 2019 A and newer for the power 9 systems we got all this work done and about a week before we were going to do CentOS turned around and went and made that announcement that I think you'll probably all be aware of we haven't quite decided what we're going to do yet we decided we'd go ahead with the upgrade of the GPU nodes and power 9 this is because most of our GPUs are actually on the power 9 that's red hat so we have support so there were didn't seem any reason to avoid that upgrade our CPU and only nodes are currently still on CentOS 7 we haven't yet decided I've put some of the options we could go with there there are plenty of others I think as well we will make a decision at some point um that's one of my you know one of our jobs for some time this year the next bit yeah MPI this is one of those areas that that caused me a bit of annoyance last year and I spent a while trying to debug Intel MPI issues I've referenced there to bug reports that are in the easy configs repository numbers there we I think we definitely experienced the first one I think that it was the second one we experienced or at least something very similar to it part way through the or part way towards the end of the year and then at beginning of this year I went 2% of our software uses Intel MPI or 2% of our installs last year uses Intel MPI I've had enough of this we're going to go I'm calling instead I using swapping out the Intel MPI for open MPI for those tool chains and that's work but basically that's because we already know how open MPI works on our cluster so we'll just use it everywhere and that seems to be going quite well and is there one of the things to note about that is most of our the software we install with the Intel Compilers in MKL are standalone bits of software where the users just use that one bit of software with not a lot of dependencies are there things like Quantum Express or NWKMR in that so you know they they benefit from the Intel Compilers in MKL or what they tend to from what we've seen but going out to open MPI is fine. Last year obviously we switched having a lot of people working from home and obvious reasons so we deployed and opened up and open on demand instance which is HPC portal to the infrastructure that we've got there to the compute nodes we provide through that that's from the Ohio Supercomputing Center we're providing a bunch of GUI apps from that ANSI Saboteurs Matlab Paraview I think there's some more that I've forgotten there also things like Jupyter Lab R Studio Server VS Code also provides shell access through a you can get into a bash shell on a compute node through a web browser and also a file browser we're finding that very useful there are alternatives to that and I know some of them have been mentioned already this week we're very much enjoying that and the last junk is I mentioned we were a tier 3 site well actually our big news of last year was the announcement of Baskerville which we are becoming a national tier 2 accelerated compute facility that's us in conjunction with the Diamondlight Source Rosalind Franklin Institute Allenturing Institute and EPSRC we're providing a tier 2 system with compute with GPUs I'll explain what we're putting in in a minute our technology partners are NVIDIA and the NOVO with OCF as our in deployment partner this is again going into our data center we're putting in compute nodes 46 one new systems one half of the one new system is two Intel Xeon processors 512 RAM and SSD the other half of the one new system is then four NVIDIA A140 GPUs apparently each of these will has about 50 53.4 teraflops per node and I forgot to copy in that from memory that's the compute sorry the GPU part and the CPUs have a bit more on top of that but I've forgotten to put down what they're rated at on here we're also 5 petabytes of storage going in spinning disks about half a petabyte in flash and one of the things we we put in last year in our own in Blue Bear we enabled the Globus endpoint which we're finding very good for data transfers and that was a precursor for the tier 2 test it on our own system first because to see whether it's all helped to provide to our partners and at that point yep I thank you for listening and are there any questions okay thank you very much Simon are there any questions for Simon either in Slack or in Zoom yes we have Yorg go ahead Yorg yes me again throwing in that question we all know about the center s problem you put it like that and you mentioned Susie you mentioned a few other distros what I haven't seen is something like Ubuntu or Debian do people actually consider that um but we probably would go down the center oh well you know sorry sorry some sort of red hat based distribution or Susie there are some questions about some of the applications we have to support as to whether they're supported on Ubuntu I'm simply thinking abacus ansis that's a decent chunk of our users um and I think at least one of those isn't supported or is only supported limited versions I can't remember exactly um I know we've had some problems setting it up on there um we might but I would think our Sus admins are more knowledgeable about red hat and prefer it so we're likely likely to go something in that line I don't know though we're still discussing we need to test it I think is the partly answer yeah I think that is really what what we need to do we need to test it on adebian based system like Ubuntu or Debian to really see can we switch over or are we really stuck in the red hat ie apm based world I don't have an answer to that as well there definitely are easy build sites who are using Ubuntu I think okra can maybe say something about that pain in the ass we all we only we only heard pain in the ass we didn't heard it part before abacus apart from that I haven't seen any problems almost anything yes unfortunately that's a big pain for us I have a chance yes um yeah it's difficult is the answer you know I think one of the things we might discuss is whether or not whether it's more work or harder work to split out in some way whether we might have bits on different OS's because we're already to some extent supporting different things but or you know deploying through some sort of container or something sitting on top there are options but we haven't explored them yet just a small comment from the easy build side if you're looking at SUSE as well that's definitely a way less popular OS than anything red hat based or anything Debian based in the easy build community and we are aware of some issues there which I think are small issues and and let's say rather obscure problems that are probably easy to fix but you may hit more especially if you're installing lots of applications if you're one of the only sites really using slas or SUSE to a big extent then yeah you may I mean you're already sort of in the corner and and that that maybe brings me to the power aspect you didn't say a lot about how how much of a pain it is to to support the power system and you definitely contributed back a lot of that stuff too easy um so we've changed focus a little bit with our power in that having tried lots of applications we now have a very good understanding of what is likely to work and which users we're targeting to use on it and what software we're going to use so we've tended to focus down the traditional AI so TensorFlow by Torch Gromax is heavily used and a couple of others yes it it's I think I personally wouldn't head towards IBM if I wanted to run everything um I would want to be focusing certain subset of applications compared to what you know I mean I talked about us having a thousand you know four thousand installs last year of most of those you know there's probably a couple of hundred of those on power in you know and a lot of that support library um one of the things we're going to red hat eight there is for CUDA versions 10 and below there is a glibc issue there is a book report on easy build with it which includes some of the links to the patches that have then been applied to some of the bits of software um but yeah seemingly anytime we change something we hit another problem so the power nine cluster you have you say you run some of the bigger applications there and and you try to upload those to the power system and free up the other system for the bioinformatics and and god knows what stuff yeah well yes so in terms of that's where our GP most of our GPUs are so if somebody's coming to us going I want to run TensorFlow I want to run PyTorch I want to run Gromax whenever that anybody asks us a question you know even if they're asking you as a completely different question we'll go well have you do you know about our power nine nodes go and use those instead or go and use those as well because that you know that's what we're finding is very good on them okay sounds good I don't see any other questions and we're getting close to the next talk so let's wrap up here thank you very much Simon


Hey everybody, it's Eric Johnson from Airtay Throws Nation and today's video we are going to talk about the barbell bent over row. Now we're going to go from a high position and we're going to talk about three big reasons why I think it's a great lift and why you should include it into your training program to improve your throwing. Now first and foremost, it's easy, doesn't require a lot of equipment and it's a pretty straightforward technical movement, but that leads into the second thing, that it really helps expose technical weaknesses in the back and the scap retractors. So if you have an athlete and they're having a hard time doing this, like the athlete I'm showing here, we're going through a lot of instruction to teach a relatively basic position. This is going to show you how significant the anterior dominance is, meaning how rounded those shoulders are. Athletes are playing video games on cell phones, they're doing so much forward movement, maybe they're doing a lot of bench pressing and they're not doing enough back work and this is really going to show you if you have a shoulder imbalance that's going to have a big impact on your throwing. So those are three main reasons why I really like it and why it should be something that you should incorporate into your training program. When we look here, we have an advanced athlete. One of the reasons why, again, I like it so much in its application is because it teaches you to keep the shoulder blades back and down and you have to pull the bar into the crease of the hip. So you're going to know that the rowing motion is more this way which pulls those scaps back and down. Because of that, we tend to use a supine grip to pull the elbows in so that we're going to get that better pulling motion that's pulling that chest open and back and this is going to be really helpful for when you're trying to improve your bench press to keep you stable on the bench. And so this is a really simple way and a good add on accessory to make sure that you're improving your bench press. When we start teaching the movement, so when you look here and I got a more advanced athlete, what we're trying to do again, like I mentioned in the last segment here, we're trying to keep the palms in a supine grip. We're going to keep the bar close to the body. So if the back is engaged, the scaps are pulled back and down, which they need to be, and then you can pull that bar kind of up the crease of the hip, like you're pulling it through your body. You're going to get those scaps constantly working down. And when you see athletes that are rounded or if you're an athlete watching this and you have those rounded shoulders, this is going to wreak havoc on your ability to create separation in your rotational throws and it's going to create a problem with you learning technique, right? Because if you have this body in a less than optimal position, you're going to start compensating. So you're going to pick up a lot of weird technical habits. So when we do this, we're pulling the shoulder blades back and down. We want to keep the shoulders retracted in position so that we're keeping that bar basically right at that top knee position. If the shoulders are dominant around forward, the bar floats out away from the athlete. So a real simple thing is to make sure that as you are doing this movement, you're pulling that bar and it's kind of sliding up the thighs into the crease of the hip. Again, the supine grip, like I mentioned in the beginning here, keeps the elbows in. So for athletes that already have that tendency to go out when you do more of a natural prone grip, meaning overhanded grip, you're going to see your athletes pulling with the shoulders and the biceps and pulling the elbows out. That's going to create an issue. So when we work here, we're going to just be keeping the scaps in place, not letting them come forward so that we can really feel that chest open up as we complete the pull. As we go through and what we want to do, whether we're looking at a beginning athlete or an advanced athlete, you're going to see that the more advanced athlete, you're going to get some bicep working. You can do this in a moderate weight, higher rep. You can do this a little heavier, but we don't want to go crazy heavy with this supine grip because you can potentially create unnecessary strain and pressure on the bicep attachment tendon and we want to avoid that. So this is going to be, we can use some relatively heavier weight, but we're going to see like our athlete here on the left, we're going to use it a moderate weight and we're going to do a lot of high rep movement so we can just really blast that back. We're going to get a little assistance work on the biceps and this is going to really be able to see where the athlete starts to fatigue and you can see our athlete here on the right. He's starting to use a little bit of a trap, but he's learning the movement and this is a really quick big bang for your buck type movement that's going to help you improve quickly. Remember, when your throws training, you want to think about what you're doing with your lifting, is it aiding in your throws development and can something simple like this make a big difference in your training? And the answer is absolutely. And this is a big part of what we talk about inside the throwing chain reaction system and our strength training for throwers segment. So if you'd like more information, be sure to click the link in the description in the bio, hit that thumbs up, give us a like, subscribe, and we will see you on the next video. As you can see, there's a lot that goes into what we do with the throwing chain reaction system. If you would like to learn more about how to structure your practices and find the things that help unlock your potential, click the link below and we will see you on the next video.


So I wanted to talk about the veil conjectures. So this is a rather deep theorem of algebraic geometry, which was first conjectured by André Veil, and was then the proof was finally completed by Deligne. And so this theorem relates the topology of complex projective varieties to the counting of points of varieties over finite fields. And this in some sense is already quite exciting because these are two different things, quite different things. And the fact that there should be any relation is kind of surprising. And so I will here not make any attempt to explain why this should be true. I will just try to explain the statement, so tell you the ingredients of it, and then give some examples. So this is somehow erected mostly in some sense the way I set it up towards the diploma students. So I will sometimes refer to the lectures that you had. So one of the advantages of this thing is that it somehow uses several things. It shows that the mathematics is a unity and that you should kind of also see it as such. So first we talk about algebraic geometry. So let me say what is a projective variety. I'm not trying to be 100% correct, so we say we work over a field. And we have the projective space, which you know. So this is pn over k, say. So these are just the one-dimensional subspaces of k to the n plus 1. And so in other words, these are the sets of n tuples a0 to an, where the ai are in k, divided by an equivalence relation. Namely, we have such an n plus 1 tuple of numbers is equivalent to the same if we multiply everything by the same constant. The lambda is an element of k without 0. So for the moment you can think that k is the complex number, so we have this. And so this is a nice space. And now we want to look at projective varieties. So for this we take the 0 set of polynomials inside this. So if f is a polynomial in n plus 1 variables, then we, which is also homogeneous, so it contains only monomials, which have all the same degrees. Then we can look at the 0 set of f, which is a subset of this projective space. This is just a set of all, you know, such equivalence classes, so a0 to an in this projective space, such that we get 0 if we replace the variables by the ai. So this is a hyper surface in Pn. And then we can also look at the common 0 set of several hyper surfaces. So if fi are polynomials, homogeneous polynomials, then we can look at the 0 set of f1 to fL, which is just the intersection of the 0 sets of the fi. So the set where all of them are 0. And so this will be some subset in this projective space.  And if they are also irreducible, so if I call this thing x, so if x is irreducible, basically means I cannot write it as a union of two smaller subsets, then it's called a right, projective right. Okay, so that's this. And so you can, if you want, you can always think that we have just the 0 set of just one polynomial, z of f. And in that case, the 0 set of f is irreducible. If f is irreducible, that means not the product of two polynomials. And you also have the notion of dimension and non-singularity, the smoothness of such varieties is defined. And, you know, there's a completely algebraic definition. So the dimension, so for instance, the dimension of the projective space will be n, the dimension of the 0 set of a polynomial inside the projective space will be n minus 1. And you have non-singularity or smoothness, which means whatever that the tangent space at every point has the same dimension as x. And for, you have an algebraic criterion for that. For instance, the 0 set of f in Pn is non-singular, if and only if. There is a, if I take the 0 set of f and all its partial derivatives is empty. There is a completely algebraic criterion, this you had, I think, in your course. So now, we want to look in particular at complex projective varieties and manifolds. In the algebraic geometry course, usually when you do algebraic geometry, you use, you have these varieties with the risky topology and this is all very strange. But if you have a variety over the complex numbers, you can also view it as a complex or differentiable manifold by just taking the same set in the projective space and using it with the usual topology. So we have that if x in Pnc. Let's say that is a smooth, so non-singular, objective variety. So maybe I could put a large n here, of dimension n. So then I claim this is a differentiable manifold, like in the course that students had. So first thing is that the projective space Pnc is a two n-dimensional, dimensional, differentiable manifold. And it's kind of obvious. If I have a point, if I have, look at the set A0 to An, such that in Pn, such that, so this was a large n, such that A0 is different from 0. So then this is a chart, diphylmorphic to C to the n, which is the same, which is for this purpose the same as R to the two n, just in the obvious way. I sent this thing to A1 divided by A0 until An divided by A0. And so we, if we do this for the other AI, we get all these charts and this makes it into a differentiable manifold. And if x is, our x is supposed to be a smooth projective variety inside this thing, then x, that's the subset x, will in a natural way is a sub-manifold of dimension. When? When n, so we always talk about real dimensions here. So, okay. So let's look at some examples. If we take one-dimensional projective space over the complex numbers, you know, one can describe this with the open set where the first coordinate, the zero coordinate is non-zero, which is isomorphic to C and we add just one point at infinity. And so this is really just the one-point complication of that. So it will turn out that this as a topological space or as a manifold is just S2. So it looks like this. And then we can also look at an elliptic curve. So let's say contained in the two-dimensional projective space should be an elliptic curve. So that means it's a non-singular variety which is a zero set of a homogeneous polynomial of degree three, non-singular cubic. Then one can show that it will be a TORS like this. Okay. And modernally, if you look again at TORS instead C in P2C is a curve, a non-singular curve of degree D, zero set of a homogeneous polynomial of degree D in three variables. Then it follows that C is the Riemann surface B where G is equal to D minus one times D minus two. So it looks somehow like this as such a handles. Okay. So we see that these are some nice manifolds. So now I want to talk about this because I want to talk about the topology of our complex projective variety viewed as a manifold like that. So, and what we want to look at of it about is the homology. The students had a course in algebraic topology where homology was introduced. So if X is the topological space, reasonable one, then we can associate to it the homology groups. Maybe I call them just H, I, X and maybe as coefficients, I take just the rational numbers. So this is a vector space over Q. And it's a bit complicated to say what the homology, so we can somehow associate to our topological space some groups or some vector spaces which somehow tell us something about the shape of X. So very roughly, we could say that the dimension of this vector space could somehow have something to do with the number of e-dimensional holes, so i-dimensional holes of X, whatever that might mean. So for instance, if we have something which consists of two kind of blocks like that, and this itself is just an open subset of two each of them, then we will find that it has two components and this will imply that H0 of this will be equal to Q squared, namely one factor for each kind of connected component. And if we have something like a circle so then we get H1 of the circle will be equal to Q corresponding to the fact that this thing has a hole in the middle which does not divide the thing but somehow does that and H0 in this case will be also Q because it's one piece and so on whatever that means but obviously this is not a definition so I will go over lectures on algebraic psychology you had some introduction of homology I think it was a simplicial homology so I will very briefly review how that goes obviously I cannot do it properly because it's a course if one does it properly so there are many different definitions for homology which are for nice spaces give the same groups but the simplest one even though not the simplest one to work with is the simplicial homology and it works very simple we divide our topological space X into simplices this is the first step so that means you have points edges triangles tetraedas and we somehow actually we will use oriented simplities so somehow so the point the point so we counter positive one can imagine an orientation of an edge to be like a direction and this will be the same as minus if you do it the other round or if you have a triangle you can view an orientation somehow as a way of turning it around and this will be the negative of turning of somehow the other direction in the course obviously you have a proper definition which somehow has to do with numbering the edges and the vertices but for these small dimensions it somehow seems to me more intuitive to just try this and then we can make so we have our X and we have divided it into simplices for instance we have this very simple case we have this thing we have divided into so we have here these are maybe our points so we have two triangles this and this maybe we give it an orientation so whatever give all of them an orientation whatever for instance like this and now we make a chain complex I don't know why it's so difficult chain complex so we have some vector space ck associated to our X we have divided into triangles like that which is just the direct sum over all sigma oriented case simplices and we take the vector space q times the simplex so we have a vector space which has dimension as many as the number of case simplices for instance c2 in this case is a two dimensional vector space and then we have a boundary map so we have now for each k we have this and we have a boundary map between different ones which is delta from ck to ck-1 which is a vector space homomorphism so q linear so the coefficients and on simplices itself it's given like that if we take a point this is map to zero if we have an edge with the two endpoints like this maybe I call this is number one and this is number two this would be and say it's oriented like this then this will be minus the starting point plus the end point and for a triangle we just take again the edges with the orientation given by this so this is say this one plus this one plus and so this gives us a map I don't tell you how it works for higher k because then you would again have to cannot rely on this intuition anymore with this little drawings and you can check if you apply delta twice get the zero map and so you can define the k-thomology group of x to be the kernel of the vector space homomorphism from pk to ck-1 divided by the image under delta of ck plus 1 by this statement this thing is a sub vector space of this one and this will be our vector space which is the homology so one has this definition it's not quite clear what this should have to do with holes but you can experiment a little bit with small cases you can somehow see where it comes from that somehow it has something to do if you have a hole because somehow it can have a k-cycle which goes so something in the zero which goes around it and you can only fill it up which then would mean that it is in the image here if the hole is not a hole so just to say it if you have a triangle like this but just the outer side then this delta of this will be zero but it will be delta of this delta of this will be zero but this is equal to delta of this okay and so therefore you can see that you will get a non-trivial group if this thing is empty but if you can close it up then you don't and so it has something to do with holes okay so much about that so once one has these things these groups we can look at the Betty numbers so b i of x just the dimension of the i-thomology group of x and we have the so we just take this number which is the dimension so we throw away some information and we have the Poincare polynomial which is just the polynomial which keeps these numbers so this is say p of x t sum over all i the i-th Betty number of x to the i so knowing this polynomial is the same as knowing all the Betty numbers so just in our examples maybe we will not explain it but for instance we get if we take the Poincare polynomial of s1 according to what I told you this is 1 plus t if we take the Poincare polynomial of s2 which is the same as you know obviously so I had to t as a variable here p of the projective space over c one dimensional t this is 1 plus t squared and if I have a surface of say genus g then we will get 1 plus 2g times t plus t squared so the dimension of the first homology group is 2g in particular for trolls we get 1 plus 2 plus t squared because g is equal to 1 so now so much for the complex things so I have explained one side of the theorem which I haven't told you and now I will tell you the other side which is somehow counting points over finite fields so for this to count points of so if we start with a complex variety if we want then to count points of a variety over a finite field we somehow have to make out of the complex variety over a finite field over the complex numbers 1 over a finite field and this we do by reduction modulo p p is a prime like 2 or 3 or 5 or 7 so but we do it so let x in complex projective space so maybe we do it like this be a complex projective, a non-singular complex projective variety as I explained before and we require that it is defined over q what does that mean so we know that x is the 0 set of some polynomials that's what it means to be a projective variety and we want that the coefficients of these polynomials are in q so if that is the case we say that x is defined over q now if it is defined over q it's in some sense also defined so over the rational numbers it's also defined over the integers because once we have coefficients in q we can multiply we can clear all the denominators of the fi and we get polynomials with coefficients in z and we can also do this in such a way that afterwards the coefficients are relatively prime ok so then clear denominators then we have that the fi are polynomials with coefficients in z as I said I want them to be relatively prime so once we have that we can reduce modulo p so we have the finite field fp so we have the fp which is just z modulo pz the simplest finite field where p is a prime number so this as a set is just the set from 0 1 until p minus 1 and then as you know you add things here by adding these numbers and then taking the rest by division mod p and you multiply them by multiplying these numbers and then taking the rest by division mod p so this is a nice field and so we can we have our fi with coefficients in z so we can take fi bar to be the so the polynomial obtained by reduced reduction modulo p so we have this has some coefficients ai and we just take for each coefficient we just take the rest that we get when we divide by p and replace it by that so we get the polynomial with coefficient in fp okay that's very simple so we have associated our complex variety with this assumption a variety over a finite field and actually fp now well we haven't yet but obviously the obvious thing is that x bar we take just the zero set of these polynomials which is a subset in the projective space over this field fp now note that okay so this is the reduction of x mod p it's a rather simple thing and now we want to so thus we have associated to a complex projective variety a variety over a finite field over this field fp and we want to count the points so note if we take this projective space pn over so so obviously we also so I write this maybe xfp we also have for if q is a prime power you know that the finite fields have as number of elements the powers of primes and these are all finite fields so if q is a power of p then this we also have the finite field so we have fq is a finite field was mentioned in the algebra course and so if you want so this is a field extension of fp it's actually a splitting field of a certain polynomial over fp so we have studied that and it has p to the n elements so anyway we have that fp is a subfield of fp to the n so therefore we can also look at the zero set of the same polynomial in fp to the n so we can also also have x bar fp to the n which is just the same zero set of the same polynomials in the projective space over fp to the n and now we want to count the points of these things so we want to count the number of points of x over fp to the n was it x of x bar over fp to the n so we should remember that the projective space of any dimension over a finite field is finite because it's a quotient of just the n plus 1 dimensional vector space over fp to the n elements and this is points of this variety are subsets so it's certainly finite and we want to count it and so we write down a generating function the generating function is a formal power series which contains all these numbers this is called the zeta function of x this would be so let's say x z our prime number of x t is a variable so what we do is we write down for all n bigger equal 1 the number of points of x over fp to the n some nice number then to make it more complicated we make a generating function like this to somehow confuse the issue even more we take the exponential so exponential just means the usual thing so the exponential of a formal power series of a variable x is just the formal power series sum n bigger equal to 0 x to the n over n factorial and as you see the thing inside here is a power series starting with t so certainly we can do this to this power series to get a formal power series this is this generating function for instance to see what this exponential does if I take I can call this exercise although it's trivial if I take the exponential of say sum n bigger equal 1 alpha to the n t to the n over n this will turn out to be 1 over 1 minus alpha times t so it's the power series which is the development of this rational function you know this just comes from the fact that this thing inside basically is the logarithm of this and so then you take the exponential so we can therefore we can use this to write down one example so what is the zeta function of p1 so I claim so where do I have it so we know that p1 over failed k can be written as k union the one point infinity so in our case over fq we have here fq and one more point so we see that the number of points of p1 over fq is equal to q plus 1 okay so if we write this zeta function here we get therefore this is the exponential of some n bigger equal to 1 so 1 plus p to the n t to the n over n now the exponential is multiplicative so it's the exponential of this first thing with the 1 and the one with the p to the n and to both of these this applies so according to the exercise this is equal to 1 over 1 minus t times 1 minus p times t that's actually relatively simple in particular we see it's a nice rational function and so now I want to say that this somehow always happens and in a very interesting way which is related to the topology of the variety we are talking about so this is the theorem very conjectures so let's say x smooth projective variety of dimension n over the complex number and so we assume that the reduction it should be defined over c defined also works anyway we will not come to that complication so defined over q and we assume that the reduction mod the prime p is also non-singular no mod p so the reduction mod p is non-singular then we have the following result first this thing this zeta function is a rational function so zp t can be written as a rational function like that so we have some polynomials with odd indices q q1 of t q of t and so on until q 2n minus 1 of t this is a polynomial and then below we have even indices q2 q0 of t times q4 of t and so on until q 2n of t where this qi are polynomials with complex coefficients in t yeah that's q2 yes they are just the even numbers there is no further complication so these are this and one can actually say one of them two of them are particular zero particular simple q0 of t is just 1 minus 1 is just 1 minus t and q 2n of t is just 1 minus q to the n you can see that for instance the case of p1 we are in this instance where we have only these two so we have only q0 and q2 and it fulfills this thing and what else can we say about these polynomials that is then more the more interesting part namely we can we are over the complex numbers we can somehow factorize these polynomials so we can write qk of t we can write this as a product over some i I will soon say more about this i 1 minus some complex number alpha k i times t so there okay where the alpha k i are some complex numbers and they have one interesting properties these are complex numbers so I can take the complex absolute value and this will be equal to p so here here I have worked over zp so everything is over p so this q becomes a p okay so this will be p to the k halves and what else can we say now comes this actually is very deep but the thing which is more that we actually can see what the index is the 0 the index goes from i equal 1 to the kth petty number of x over the complex numbers so this is very amazing so that means the degree of this polynomial qk of t is equal to the kth petty number of xc and this is really quite amazing because it doesn't look like there's very much relation between counting these points and the topology of this thing okay and the last thing which I'm not sure we will use is that there's actually some relation between these different polynomials namely if I take q2n minus k it goes on to n of t this can be written as the product again i equals 1 to pk of p so 1 minus p to the n times alpha k i so this is something which is related to probability for any faults I don't think you know that but anyway there is a further property that this thing has but you can see this in particular as you see implies that pk and p2n minus k are equal and this is quite a equality which one knows from algebraic topology but maybe not in a beginning course like that okay so what one can easily check that one can reformulate this in a simpler way so this formula so this formula 1 and 2 we can reformulate in a different way so 1 and 2 are equivalent to the following if we just take the number of x over x bar over fp to the n then we can write this as the sum k from 0 to 2n sum i equals 1 to the i th petty number of the our complex variety minus 1 k times alpha k i to the n where the alpha k i are the same as we had here so this is just if you think of it it's just again applying this exercise it's really an exercise that these two statements are equivalent but what this tells us however is that if we count the number of points over these fp to the n for all of them then we can determine these numbers I mean you can determine the terms in this formula and we can read off the petty numbers we just have to see when we know all these numbers we know what the alpha i are and we have to count how many alpha i have absolute value p p to the power k half this is the case petty number of our variety okay so this can read off the petty numbers or if you want the pancrepolynomial of our complex variety from the number of points over fp to the n if we know it for sufficiently many n and in particular in particular a simple case so if we are in the case that this is very very simple that the alpha i are just powers of p so if number of x over say f so q is p to the n is a polynomial say g of q then it follows that the pancrepolynomial of xc you know we just we have to count the things where we have the power p to the k half so we just replace q by t squared but in general you know we don't need that so even if it's not like that we can do things okay so i want to give some examples so i should say so for one thing it's exciting that this action is there between complex geometry and counting points but it's also actually useful it has been applied so the field of algebraic geometry i'm working in is modular spaces so modular spaces are varieties which parametrize something we are interested in so the points of the thing corresponds to certain objects say vector bundles curves whatever so then usually you can prove that such a modular space exists and maybe you can prove it's non-singular and projective but maybe you cannot say very much about its shape but you know what its points are so therefore you can count the points so without knowing anything more than the existence and the kind of smoothness you can compute the homology of the space we have no idea how it looks like but it is based by just counting points so and so let's look at some examples of such modular spaces in some sense for instance if we take the n-dimensional projective space so if we take this over fp fq so this is just parametrizes the one-dimensional subspaces of fq to the n plus 1 and so we use a basis vector so this is fq to the n plus 1 we take a non-zero vector and we divide by multiplying by constant by so we have the so we have equivalence classes no n tuples non-zero divided by by this and so this means if we take the number of points this will be n plus 1 minus 1 points and this has q minus 1 so we get q to the n plus q to the n minus 1 plus and so on plus 1 so we get this so this is a polynomial in q so according to the story we get that if we take the Poincare polynomial of pn this is just say whatever t to the 2n plus t to the 2n minus 2 plus and so on plus 1 and in fact it's kind of well known that the homology of p2 has one is z in every even degree and so this is over the complex numbers and so we get this let's look at this somewhat more complicated but related example which is the Grasminian so this is a projective variety which parameterizes so g I call g ln so this will so over any field this will parameterize the l-dimensional subspaces of k to the n so the projective space is just g1 n plus 1 how can we count points so we want to compute the number over a finite field fq so how does it go so if we want such a subvector space so it's a subvector space we first can choose a basis for the subvector space d of dimension l so we choose just l linearly independent vectors how does it go so the first vector we can just choose any nonzero vector so we get q to the n minus 1 so this is v1 is an element in k to the n 0 then the second vector we can take any vector which does not lie in the span of this so this would be q to the n minus q so the v2 is an element in k to the n minus the span of the first vector and obviously you can see how it goes on so we get in the end q to the n q to the l minus 1 because our last vector vl is an element in k to the n so k was fq without the span of the first l minus 1 which is in l minus 1 dimension of vector space and then we have to forget the basis so we just want to remember the vector space so if we have two bases of the same vector space v1 to the l depends the same space as w1 to wl they are both bases of a l dimensional sub vector space then this is like this if and only if there exists a matrix which is also then unique and l by l matrix which kind of transforms one basis into the other so such that wi is equal to sum over all j m aij times vj for all i from 1 to l that's how it goes so we can change the basis so we have to count how many such matrices there are but if you think of it the first in l matrix of rank l obviously as opposed to it should be matrix of full rank if you take the first column of the matrix it's just a non-zero vector from k to the l the second one is a vector from k to the l from the compliment in k to the l of the one dimensional sub vector space the first one and so on so it's the same story as here so the number here matrices is q to the l minus 1 times q to the l minus q times and so on q to the l minus q to the l minus and so we find that the number of points of g q is what we obtain if we divide this thing by that and we can kind of factor out the extra factors q that we have here so we will get that this is that this is just anyway it seems obvious q to the n minus 1 times q to the n minus 1 minus q to the n minus l plus 1 minus 1 divided by q to the l minus 1 q to the l minus 1 minus until q minus 1 and so thus it means if we want to see what is the Poincare polynomials what are the Betty numbers the Poincare polynomial of which we don't know anything else than that it exists and that parameterizes these things and I claim it's also non-singular and so on the Poincare polynomial of this thing over the complex numbers will be this thing where we replace the q by t squared so this is t to the 2n minus 1 and then it goes on until 2n minus l plus 1 1 divided by 2l minus 1 and it goes on in the same ways here until t squared minus 1 and if you are familiar with some combinatorics and so it's maybe not completely obvious that this is a polynomial but you can easily check that it's one factor by the other and we know that by this thing the odd Betty numbers of this thing are 0 and the even ones are given by this formula so if I take the 2k Betty number of this thing you can read off with some effort from this formula I will not even exercise that this will be equal to the number of partitions of our number k into if I'm not mistaken n at most n minus l parts and all of which are smaller each of which is smaller equal to l so you partition means just you write the number k as a sum of smaller numbers and I require that I should have at most n minus l numbers and each number should be at most l and for instance the first non-trivial case is the g2 4 so when it's not a projective space in this case we get the primary polynomial of this thing will be t8 plus t6 plus 2t to the 4 t so I used up my time I had some I wanted to say also something else but maybe I will not so ok so maybe that was all I wanted to tell you for now thank you very much I hear you no I mean what no you see when I do this thing I do the reduction modulo p inside this finite field but there's nothing to do the real parts the points over the finite field do not lie in this other one because you reduce modulo p so it's not you cannot see it in this way ok I said it in the beginning so the conjecture was maybe made in the 40s and it was proven I think 74 or something by the proof was finished by the linear 74 it got the fields medal for that there had been work on that for many years by Gotendieg and others to set the basis but it has been proven for some time somehow it's always called the veil conjectures because it's because it's not the veil theorem because he didn't prove it he proved it for curves and then conjectured it in general no no I mean I was I understand that I was trying to say it in a simpler way so the correct word is good reduction mod p I just said pretended that meant it's just smooth but I somehow yeah okay but I don't know it's maybe a bit complicated to say what good reduction precisely means but it roughly means that it's non-singular certainly it must be non-singular of the correct dimension and maybe it should be irreducible and whatever but anyway so it's I require that the model but you know if you have a natural so so this is just for simplicity so the whole thing works also when it's not defined over q over the rationales it's a bit more complicated so if your variety is not defined over q but just like that over c you have some polynomials then you can look at the field which is generated by the coefficients of the polynomial so this is a finitely generated extension of the rationales and then you can you can look there you can find some residue field which is a finite field of characteristic p and then you work with that so to do it just that I can work with fp like this and just reduce mod p for this I need to that is defined over q but I don't actually practically need it but obviously when you have a naturally defined generalized space most of the time it will be defined over a rather small field because it's naturally defined for instance like if you take the Hilbert scheme of points it will be defined more or less over the same field over which the variety was defined and so then obviously you cannot make such assumptions and you have to do this thing you take the field generated by the coefficients and then you reduce you find a finite field which is a residue field of that well if you do it in a suitable sense quasi-projective is enough maybe one has to know what one means by the Betty numbers then I think you might be getting the Borel Moore homology or something like that but it certainly makes also sense then in fact well maybe I will not say that


So here's the Nubotan ARM Cortex M23 solution right here. And hi, so who are you? I'm Robert Lin, I'm responsible for the M251, the latest ARM VAM secure IOT microcontroller in Nimotor. So you are the first or one of the first to use ARM VAM right? Yes, and we just got both PSA and PSA functional API certified. So you have this here it says PSA certified trophy right here. So that means you have full security and the microcontroller right? Yes, and this microcontroller targeting secure IOT applications. So what is this board? What is this? This board is actually a maker board we called, but it's not just only for evaluation. It can be easily to connect it to IOT cloud. For example ARM Pellion system. So right here you're talking about it. What is the text here? What do you say? Pellion secure IOT is based on Pellion cloud. Pellion cloud. What is a Pellion cloud? Pellion system is endorsed by ARM. A convenient secure remote IOT operating system. So here it says the tree. It says you have a Cortex M0 M4 M23 and you have 8051 also and ARM 9. Correct. So what is a company where you based? We based in Taiwan. Yeah. And the CPU cores represented our product portfolio. So you can see we cover a lot of CPU cores and we try to maximize these CPU cores to develop many MCU products for customer, for application requirement. And what is the other solution you have around here? For example this one right here, what is this? This is an industrial remote terminal unit based on our... This is an industrial remote terminal unit. It uses our ARM 9 series products. It runs Linux? It can run Linux. To do what? This is a gateway like a product. So it can operate with a gateway and to receive many different communication protocol or data. Nice. And is this using the M23 Yes. So you can do a secure fingerprint unlock. But it doesn't just be limited on secure fingerprint. It can also offer more resources for enterprise development. So how did you make the M23 so quickly? How did you work on this? How long time did it take? Yes. We invested on our VAM focus on M23 three years ago. Three years ago. Yes. At the beginning. Yes. At the very beginning. We are one of the first wave partner with ARM for this new CPU core. Is this shipping mass production now? Sure. Since when? Sure. When is the last September? September 2018. Yes. Mass production. And this is the idea of the ARM V8M, right? You have a security? Yes. Secure element or secure... We should say trust zone. Trust zone. Architecture. Today many people are still confused or struggled to use trust zone MCU because it's different from traditional secure element. But we try to develop trust zone applications as the references for the market. Then people can quickly catch up with this concept and develop a useful IoT application based on trust zone MCU. It's a big change for the embedded world, right? Right. Is it a big change? Right. Because other people do security without the trust zone but it's not the same? Yes. We added a lot of trust zone beyond the features. So customers can use these features cooperating with trust zone securely, conveniently to protect their software assets also remote network communications. This is a big revolution. Big revolution. Yes. And around here you have some more solutions right here. So there you're seeing human machine interface solutions. What is this? Maybe Tyler can help you. Hi. I'm Tyler from Dimvaton. I'm sales from Dimvaton. Tyler. So what is this? Oh, so okay. So right here is our M4 solutions. This is an M4-4. The same as this? No, this is an R9. Oh, R9, yeah. Yeah, this is an M4-4. They connected to a camera CMOS sensor and it reads a serial number over here. So you can see the M4-4 can drive a display over here. And is this sending the signal? Yes, it's sending the signal to Bluetooth to this M4-4. This is also another M4 inside. It drives an LED. So it reads the number, it transfer it to it and then if it's in real time it runs some kind of neural network machine learning algorithms. So it used the machine learning algorithm to do the pattern recognition. So if you can see right over here, I changed the number here. I put the three in the front. And as you can see it's lined up in here. So there's a potency press here. Yeah. Press it. As you can see it... Update. Yeah, it reads the number and transfers it to it. Is this an M4 also? This is an R9. R9 solution. So this is a human interface and you can do something like touch screen or for as you can see there is a telemetry data. Yeah, telemetry data over here. So if you read some kind of data from the sensors and you can transfer the data to something like a screen, so you can read it easily. Also we have license from SEGGER, the EMWIN library, so you can do the GUI very quickly. Nice. If you use our chip, R9 chip, you are using the EMWIN library for free. Nice. And what do you have over there? Okay, please. So right over here. Can you stand behind there? So what is this? This is an Dolly 2 lighting control system. It's a protocol by Philips and as you can see it's connected to multiple controller boards with applications like this. The one is for data and one for power. It's not mesh. It's not mesh. It's not mesh. So how does it work? So they use the differential wire, then we call it the Dolly bus. Dolly bus. Letting control. So may I show you? If I choose, now if you want to control this one, you can switch the toggle and change its color. So what is the market for this? Market for this like home control automation or light control in your household or if something like you want to do something. Isn't it a chain? It's in a chain, but you can do separate control. Nice. So each one has an ID and they can tell which one to do what? Yeah, as you can see it's fine. Like in the board, you can control every light. All the lights. All right. You can control every light and also that we support multi-master mode. So that for multi-master means that you can control independent light. So this master doesn't work and you can automatically switch them on the master. Hi, so who are you? Oh, my name is Jenny. I'm the marketing from Novotan. So as Novotan has a big market? Yes, we are a quite global global company and we basically are our headquarter is in Taiwan, but we have the office globally like in China, in US, in Israel, and even in India. The company is how old? It's 11 years old. And it's always doing IoT embedded stuff since the beginning? Since the beginning we are developing our microcontroller and now we try to expand our platform to like the IoT gateway or IoT security. And how many engineers? Is it a secret? No, the total personnel is around 1,500. 1,500? 1,500. In Taipei? No, in Xinzhou. But we also have a sales office in Taipei. And what are all these boards? Are there different solutions you have? Yes, this is a different one. Like this is an ARM9 solution. You can connect it to the zero port. And there's a different RS module that can combine with the Arduino. What is this? You said there's like the narrowband IoT or Wi-Fi module and also like Lorawen, different kind of the four kinds of RS modules. Which one is the most popular solution you have? I think that everyone is very, everything is very popular. Everything is very popular. All right. But in the future are you going to do any Cortex M33? You have any announcement about this or? We are still planning for it. Cannot have it exclusive for my video. Okay, cool. All right. But thank you very much.


The Taliban is a complex religious and political movement. The Taliban initially emerged in 1994 in the southern province of Kandahar. This is a time when Afghanistan has disintegrated into anarchy and civil war following the collapse of the last cross-Soviet government of Reza Najibullah in Kabul. The Taliban emerged during this period with the slogan of bringing peace to Afghanistan. And its emergence was facilitated by critical support it received from Pakistan, especially Pakistan's military intelligence, the ISI. Initially, according to most reports, it began with a really humble beginning locally. It was led by Mullah Muhammad Umar who was a local religious figure but with no known reputation in Islamic law, jurisprudence or Afghanistan's national politics. He and his group of students first started disarming locally abusive militias and then they were also starting supporting Pakistan's ambition at that time in 1994 in Afghanistan to open a trade route through Afghanistan, through the cities of Kandahar and Herat to Central Asia. So for the first time we also see that the Taliban emerged as a force that achieves this really standing victories against this locally abusive militias that were fighting for control of the city in the province of Kandahar. And then subsequently as they seized control of territories and provinces across Afghanistan, they reached Kabul in 1995 and they became just another important player in Afghanistan's bloody conflict for the second half of the 1990s. They started implementing the version of the Sharia law as they interpreted which was really a totalitarian interpretation of Sharia which they would use to impose their will on every aspect of the Afghan society. The Taliban are remembered now with executions of women and other members of the society and public places like stadiums and they also engage in mass killings of the Shia Hazaras, one of the most important of which is August 1998. After they took control of the city, they went on a campaign of killing primarily the Hazaras residents of that city. After they achieved control of the northern city of Mazar-e-Sharif in August of 1998, three countries extended diplomatic recognition to the Taliban. This one Pakistan, Saudi Arabia and United Arab Emirates. I live part of those years in Afghanistan and I witnessed firsthand how the Taliban ruled the country. They turned the iPhone radio network into a mouthpiece of propaganda calling it Sharia. Afghanistan of the Taliban from 1996 to 2001 is one in which there is no video, no photograph, no art. They destroyed famously the Buddha statues of Bamiyan, such a monumental important heritage of Afghanistan's civilizational past. By the end of 2001, following the US military intervention, the Taliban regime, the Taliban as a movement disintegrated. And that explains also in important ways how vulnerable, how fragile the movement was because it did not enjoy a lot of popular legitimacy in Afghanistan. Another important layer of the Taliban would be described as local Taliban. These are people who have grievances at the local level, especially in the context of the post-2001 Afghanistan, where sub-national governments proved to be one of the main failures of the international intervention. It created a lot of opportunities for abuse of power by local power holders and as a result there were many people who were pushed into the heart of the Taliban as a result of those governance failures, abuse of power at the local level. And then at the top of all of this, I think we should also keep in mind that the Taliban have also been evolving over the past few years and months in particular. There are a range of other groups that have joined the Taliban for pragmatic reasons, primarily seeking to ensure their survival in rapidly changing security environments in Afghanistan. This is especially true of the past few weeks and months. I suspect there are many groups locally who switched sides to the Taliban just to ensure their survival. So just to sum it up, I think it's important to keep in mind that the Taliban is a very complex movement. It was best described by an analyst of Afghanistan, Thomas Rottick, as a network of networks. The impact of the US intervention on the Taliban has been profound. I think that for the first time the Taliban found itself in a fight with a very major technologically advanced power like the United States. So as a result of this, to fight back, they also needed to adapt to the changing circumstances. And I think one of those really important adaptations by the Taliban is their appropriation of very effective public relations campaign propaganda. So far what I'm seeing is the propaganda wing of the Taliban having quite effective in telling the world what it hears at the moment. There are differences of views and different groups in that system represents the diverse groups of Afghanistan. For me, I would like to be pessimistic unless I see evidence is coming that proved me wrong and I would be quite happy to see that.


All right, folks, I will preface this conversation by saying I am well aware of the fact that it is way too early to begin speculation about the 2024 presidential election, given that we haven't even seen what happened with the 2022 midterms yet. Having said that, though, it is looking more and more likely that we're just going to see a rehash of 2020. We're going to see Joe Biden versus Donald Trump again, because Joe Biden is looking more and more likely to want to seek re-election. And let me just remind you that he will be 82 years old in 2024. So as Brett Samuels of The Hill explains, President Biden intends to run for re-election in 2024. White House Press Secretary Jen Psaki confirmed Monday following a flurry of reports on discussions among Democrats about whether he would follow through on a bid for a second term. That's his intention, Psaki told reporters, confirming reports that Biden has been telling advisors and staff he plans to run again in 2024. The Washington Post reported over the weekend that Biden and his closest advisors have been reassuring allies in recent days that he plans to seek a second term. That report followed a political piece last week about a positioning among other Democrats who may be interested in running for president should Biden decline to run for re-election. While Biden's plans for 2024 could change over the next two years, sources close to the president have maintained, since he took office in January, that he would campaign for a second term. So I feel torn about this, not necessarily out of preference, but strategically speaking and whether or not this is going to make any difference in terms of the Democratic Party's ability to defeat Donald Trump, assuming he is the Republican Party's nominee. I think that if he steps down and allows someone like Kamala Harris or Pete Buttigieg to run in his place, do they have a better chance than Joe Biden at beating Donald Trump? I actually think they have a worse chance against Donald Trump. I mean, I don't think that Joe Biden would fare well against Donald Trump, assuming there's going to be a round two, because I think that COVID-19 really changed the dynamic of the 2020 election. I think that Trump was the favorite. But because of COVID, Joe Biden had that advantage because Trump bungled it so much. But come 2024, COVID-19 might not be as prevalent of an issue. We don't know what that's going to look like. People are going to see that Democrats didn't actually deliver, assuming they didn't deliver and they'll forget how bad Republicans were. And they'll just choose to stay home and that will pave the way to a Trump victory. So either way, I think that the conclusion is Democrats are fucked. I mean, really, if they run Kamala Harris or Pete Buttigieg, does anyone actually believe that they could fare better against Trump?  Their policies are milk toast, and perhaps they're to the left of Biden, at least aesthetically speaking, but they can't excite the base. The reason why the Democratic Party came out for Joe Biden, even if they disagreed with him politically, they supported Medicare for all, so they agreed with Bernie in theory, but they still voted for Biden was because of electability. So if Donald Trump is running again and assuming there's widespread fear among the Democratic Party's base, that Trump can be elected once again, well, then do you think that they're going to opt for someone who they think can be the future of the Democratic Party or do you think they're going to be more inclined to stick with what they know? And that's Biden, someone who already proved that he can beat Donald Trump other than rather than, I should say, an unknown quantity like Pete Buttigieg and Kamala Harris. And just like from my perspective as a leftist, I honestly don't care. I'm not invested in a Democratic Party primary.  But the Democratic Party, I mean, who's there to be excited for? There's no progressive leader that is going to run in 2024, at least from this time in 2021. It doesn't seem like anyone's going to run. Bernie's not going to run again and he shouldn't want to run again. I mean, he should take a break. He tried it twice. He gave it his best. If I were Bernie, I wouldn't want to run again. But I mean, who's going to run? Besides a corporate Democrat, the most progressive that we get is going to be who like Stacey Abrams. I mean, that's just they're all the same to me. Honestly, all of the corporate Democrats to me, they all blend together and that's bad because ideologically speaking, you want diversity, you want to see a robust Democratic Party primary. But in the event, let's assume that Biden didn't want to run again. And we saw a Democratic Party primary. I mean, what's there to be excited about? Another better or work versus people to judge Obama impersonation contest. I mean, the Democratic Party, their prospects right now are terrible. They have a lot of people who are prominent figures. Who are terrible.  Like, I think it's bad that Joe Biden wants to run again because he's bad. But then again, the other options are bad as well. So damned if you did, damned if you don't, they're fucked either way. So I don't I don't even know what to say. If we do see a rehash of Biden versus Trump, I will say that that really speaks to how flawed our system is and how we definitely need electoral reform so we don't keep saying seeing the same fucking people again and again and again like we need to move past the Clinton Obama era and move to a different era. But I'm afraid that the next era is going to be as shitty as the current era because you're not really changing the structural flaws that make the Democratic Party shitty in the first place. I mean, it is a party that has been captured by special interests and large multinational corporations and to the extent that they want to implement changes, there's always going to be at least one or two figures that holds the entire party back like mansion and cinema. So people are feeling rightfully demoralized. And I think that the Democratic Party isn't helping with that. But having said that, though, Biden is going to run again in 2024. We need new leaders in this country. I think that most of you can agree with that. But is there anyone remotely exciting, anyone who is at least somewhat progressive and not a fake progressive? I don't know. But seeing this. It depressed me. I'll be honest, but it's way too early to get depressed and down about 2024. I'll focus on 2020 zoo and all the demeritism and gloom that comes with that. The inevitable bloodbath and return of Republicans to power.  Why do I talk about this shit?