Thanks for the invitation and the introduction. So the title is quite a mouthful, and the talk that I will give will be about joint work. And let me tell you who my co-authors are. So it's a joint with Jonas Bergström and Adrian Diakony and Craig Westerland. And there is a second paper, which is also a joint with Jeremy Miller and Peter Pappst and Oscar Randall Williams. And both papers are on the archive, and I will try to make some sort of distinction of which results I'll talk about are from first and which are from the second paper. So the subject of these two papers is moments of the Emily of Craddick L functions over a function field. So let me describe the problem. So we'll fix Q, sorry, an odd prime power, and we consider the following sum, with sum over polynomials of degree 2g plus 1, and we want these to be monic and we want them to be square free. And for each such polynomial, one can consider a hyperliptic curve with equation y squared equals d of x. And the condition that it's square free says that this is a smooth algebraic curve over this finite field. And as such, it has an attached L function. So let me call this curve cd. And what you can do is you can look at the zeta function of this variety and you look at the numerator of the zeta function and that's some polynomial evaluated at Q to the minus s. And this is of degree 2g. And the roots of this polynomial are exactly the eigenvalues of Frobenius on the first comodic group of this curve etal comodic group. And this numerator of the zeta function defines the L function. So we define this to be L of s chi d. And what we're going to be summing is the central value of this L function. So there's a functional equation that relates the value at s and at 1 minus s. And 1 half is the central value and we raise this quantity to the rth power. And finally, we want to normalize by dividing by the number of terms in this sum. Actually, it's going to be more convenient to just normalize this by Q to the minus 2g minus 1 which is not quite the number of square free monic polynomials but they differ by an easily understood factor. So we want to understand this quantity as the degree of the polynomial goes to infinity meaning the genus of the curve. Let's scroll down a bit. And the conjecture is... Let me give this quantity a name. I'm going to call this mr of g. And the conjecture that comes from Conry Farmer Keating Rubenstein-Snaith is that this quantity as g goes to infinity is given by an explicit polynomial plus an error term that goes to zero where u r is an explicit polynomial degree r times r plus 1, 2. I'm not going to tell you the formula for this explicit polynomial. It's not very enlightening when you see it. And it's a massive formula. But the content of the conjecture is not a specific formula but it's rather the recipe that they give for how to predict what are the moments of various families of L functions. So this is an instance of a family of L functions that we're interested in. And we are looking here at the arth moment of the central value of this L function. So this is just one special case of a very general method for giving predictions for what are these moments. And it's very refined information because it's not just a dominant term but all of the lower order terms. And before I go any further I should make some comments about this conjecture. The first comment is that you can derive the same conjecture by different means. And this was done by Diakonio Goldfeld and Hofstein using multiple Dirichlet series. So they gave a different way of obtaining conjectural asymptotics using conjectured analytic properties of these multiple Dirichlet series. And it gives the same conjecture. And I should say that in the original paper of Conry, Farmer, Keating, Rubenstein, Snaith they do look at the quadratic family but not the quadratic family over function fields but over number fields which would be quadratic Dirichlet L functions instead of these L functions attached to hyperliptic curves. I didn't say the word hyperliptic curve but hyperliptic curve means a curve that can be written in this form y squared is d of x. And the general recipe from Conry, Farmer, Keating, Rubenstein, Snaith was carried out in a different paper of untroud and Keating. So they worked out what should be the prediction in the function field case. And this philosophy of moments goes back to study of the Riemann's data functions. So in the case of the Riemann's data you could be interested in there's only one Riemann's data function so it doesn't really make sense to fit it in a family but what does make sense is you can study the distribution of the value along the critical line as opposed to the value at one half. So in a sense that's a continuous family of L functions. And if you're interested in the value distribution of zeta along the critical line you can study the moments like this. And a conjecture which is wide open is that it should be asymptotically equal to C, R, T, log R squared T. Where C, R is an explicit constant which was conjectured by Keating and Snaith using random matrix theory. And this recipe of Conry, Farmer, Keating, Rubenstein and Snaith can be used to give a more precise version of this conjecture which doesn't just give the dominant term but all lower order terms as well when this is given as a polynomial in the logarithm. And this conjecture is known for R equals one and two goes back to Ingham and Hardy and Littlewood but beyond that it's not known. But their original motivation was the study of the Lindelöf hypothesis. The Lindelöf hypothesis can be formulated equivalently in terms of these moments of Riemann's zeta function and that this should be bounded by T to the one plus epsilon for all big epsilon bigger than zero. Now for the quadratic case maybe I should label this conjecture star. Star is known for R up to three and partially R equals four. By partially I mean that one knows the first q coefficients of the polynomial but not the whole polynomial. This is largely due to the work of Alexandra Florian. In the number field case the analogous question for the quadratic family is also known for R up to three and the R equals three case is due to Sander Arrajan. And you might think that the number field case should be harder than the function field case because in the function field case we know the Riemann hypothesis and we know a lot more than the Riemann hypothesis. We know the Riemann hypothesis for we know the Riemann hypothesis so to speak. But previously it had seemed that the number field and the function field case were on the same order of difficulty because we are not much further in the function field case but in fact it turns out that you can go further in the function field case. And the theorem that comes out of the two papers one with Bergström, Diakony and Westerlund and the other with Miller, Pazzt and Randall Williams is that this quantity MR of G you can write this as this explicit polynomial QR of 2G plus 1 plus an error term that looks like big O of 4 to the power G times R plus 1 Q to the power minus G plus 6 divided by 12 and the point is that if Q is sufficiently large this is indeed an error term so 4 to the G times R plus 1 it grows very fast with G but the other term Q to the minus G plus 6 over 12 goes to 0 exponentially fast and if Q is sufficiently large with respect to R then this is indeed an error term fixed R once we choose Q sufficiently big this conjecture star is indeed proof that's a statement of the theorem are there questions at this point I'm going to say something about how we prove this and as I hinted in the title of the talk this is proven using homotopy theory I just feel like I don't have a clock am I sharing this so here we go so the method is to use homotopy theory and the link between topology and arithmetic is via the growth and declifted trace formula so I indicated at the start of the talk that we know a lot more than just the Riemann hypothesis we know the grand Riemann hypothesis in a sense we know even more than that because we have a full formalism of etal homology and etal sheaves and 6 functors formalism and whatnot and this growth and declifted trace formalize part of this thing will be on the Riemann hypothesis let me state the growth and declifted trace formula if you have a smooth scheme over fq of finite type v is an L addict local system on x and there's a formula that looks like this you sum over the fq points of x and you take the trace of the Frobenius on the stock of the local system at x and this is the same as due to the dimension of x times the trace keep writing here Frobenius of the homology of x with coefficients in b this is etal homology and I should make some comments on this formula if you know the v-conjectures that concerns the case when v is the trivial local system and when v is the trivial local system then on the left-hand side you're just taking the trace of the identity on the one-dimensional vector space and that's o1 so you're just counting the number of points of x over fq and the right-hand side is what gives a homological interpretation to the number of points on the variety but in this form you can also allow for a weighted account of the number of points and I assumed that x is smooth here in order to write it in this particular form which will be the way it's used in the argument if it's not smooth you can write it in terms of compact support homology but for smooth schemes compact support homology is isomorphic to homology with a tate twist and that tate twist explains why you take q to the dimension of x because that's the amount that you need to take twist and the trace on the right-hand side is taken in a graded sense so the homology is a graded vector space and what it really means is the sum of the traces and all of the even degrees minus the traces and all of the odd degrees so it's a usual super trace and so far it doesn't really link arithmetic to topology because both sides here are very much arithmetic objects but the further result is the art in comparison theorem which says that et al homology of x with coefficients in v can be identified with a singular homology with a set of complex points of x and the analytification of this local system and there's a canonical such identification here for the art in comparison theorem to be true I need x to be a subfield of the complex numbers so certainly not a finite something that's defined over a finite field so to make sense of what I'm about to say you can assume that x is defined over the integers and then if you have x defined over the integers you can both embed it into the complex numbers you can also reduce it modulo prime to get something defined over a finite field for prime power and then you're forced to ask whether the et al homology of your reduction mod p coincides with the et al homology of the thing that you get by embedding into the complex numbers that's not always true it's true for generic specialization at some prime in our setting it's going to be true at every prime and that's because the schemes we're looking at have normal crossing divisors that are smooth normal crossing divisor compactifications that are smooth over the integers but this is all technicalities the point is that for all the schemes that we care about the et al homology which is the thing that controls the number of fq points and these weighted counts the number of points can be identified with the homology of a corresponding complex algebraic variety that can then be started by topological methods and in our case we're going to take x to be the configuration space of n distinct ordered points on the affine line so this is smooth scheme over the integers and it makes sense to reduce it modular prime and it makes sense to embed it into the complex numbers and if I form this embedding into the complex numbers I get the usual space of n distinct ordered points in the plane which is a k pi 1 and it's k pi 1 for the braid group on n strands beta n is the art in braid group and this is good because if we go back to the original conjecture which is up here we were summing over monic square free polynomials of a given degree and that's exactly the same as summing over fq points of this configuration space of distinct unordered points because monic square free polynomial can be uniquely identified with a configuration of distinct unordered points namely its roots and in the growth and declifted trace formula we were summing over the fq points so it's going to turn out that the right-hand side of the growth and declifted trace formula becomes the right-hand side of the conriform or Keating-Rubinstein-Snave conjecture and the left-hand side is also identified and the particular local system we care about is going to be of the form of the exterior algebra on some other local system w where w is the reduced borough representation of the braid group specialized at t is minus 1 I take this representation of the braid group called the reduced borough representation that depends on the parameter if I specialize it at the root of unity the monodromy of this representation is related to the family of curves y to the p equals d of x so if I take a primitive pth root of unity I would get something related to that particular family of curves and when I take t is minus 1 I get exactly the monodromy of the universal family of hyperliptic curves and I should take a half integer t twist of w to be correct and the description I just gave of w is only valid over the complex numbers but it turns out that there's a natural way to define it in an analytic setting and it's what you should do and the upshot is that we need to estimate the trace of the Frobenius on the homology of the braid group the coefficients in this thing I guess I should do quite this differently what you're taking the trace of depends on r r is the moment and it depends on r in this way and why is that the left-hand side of Groten-Dick-Lefchitz trace formula gives the left-hand side of the CFK or S conjecture the right-hand side of Groten-Dick-Lefchitz is exactly this trace of Frobenius up to this factor q to the dimension of x the dimension of x is n and q to the n is exactly this normalizing factor that I divided by formulating the theorem or the conjecture and the estimate is as n goes to infinity and if you are if you come to this talk from a topology background which maybe not a lot of people are doing this immediately suggests that what you should be doing here is something about homological stability and homological stability is paradigm that's been important in topology since the 70s and applies to many families of discrete groups that are interesting to topologists let me say a few words about homological stability so you can think about topological groups or more other general families of space but let me formulate things for discrete groups suppose I have a sequence of groups like so of the braid groups or the symmetry groups or general linear groups over some ring these are all very good examples to keep in mind and you can ask if these satisfy homological stability and this would mean that these homomorphisms or isomorphisms or k less than say some constant times n then we say that the sequence has homological stability and what's interesting is that this happens for many well studied and natural families of discrete groups such as symmetry groups, braid groups, general linear groups over some rings mapping class groups moreover computing the homology of the groups individually in the sequence rapidly becomes intractable and also maybe the answer is not so structural whereas the stable homology which is the limiting value that you get when you send n to infinity is more computable and in many cases more interesting reveals new structure and I formulated this here just as homology of the groups themselves which I guess implicitly means constant coefficients but it's also interesting to study homological stability with twisted coefficients this has also been done in these classical families of groups like the symmetric groups etc etc and here we're looking at the homology of the braid groups and we're taking the limit of the homology in some sense as n goes to infinity and we're taking a very nice family of coefficient systems for which it was in fact known that homological stability would hold already before our work and it seems that what we should be looking at is what is the stable homology of the braid groups with these particular coefficients already before us it was known that braid groups have homological stability for any polynomial coefficient system I'm not going to tell you precisely what a polynomial coefficient system is and all of these exterior powers of this Bura representation is polynomial this exterior power wedge K of W is polynomial of degree K and W itself is linear polynomial of degree 1 which is good however the stable range is of the form this map is an isomorphism for K less than n over 2 minus the degree of this coefficient system and this is come from a paper of Oscar Randall Williams and they give a very general homological stability theorem that applies to many different families of discrete groups and it always gives for all these families of groups a stability theorem for polynomial coefficient systems and it always has this shape where it depends on the degree of polynomiality and now there is a problem that arises which is that here we were looking at the full exterior algorithm on W which is the direct sum of all exterior powers and these include summands of arbitrarily high degree but that's bad in this theorem because then the degree is unbounded and the effective range of homological stability is zero when you're taking homology with coefficients in the full exterior algorithm which is the thing that we want to understand so something different must be done somehow in our case where V is the full exterior algorithm on W the degree is unbounded and moreover even if we did know homological stability for these coefficients meaning the full exterior algorithm W that would give us the wrong answer because if we had stability for this full exterior algorithm to answer with itself r times then we would end up proving that this quantity of r of G converges as G goes to infinity to the trace of Frobenius on the stable homology with these particular coefficients but this is not what we wanted to prove we wanted to prove that this asymptotically grows like a polynomial we don't want to show that it converges to some number so we don't Randall Williams-Valls result we don't want to prove stability and we also don't expect stability to be true because it contradicts the CFKRS conductor so something different must be done and what is better to do is to decompose this full exterior algebra into irreducible representations for the symplectic group and where did the symplectic group come from well this Bura representation is actually a homomorphism from beta 2G plus 1 to SP 2G Z so this representation W goes from the braid group to the symplectic group and then if I decompose the exterior algebra on the fundamental representation of the symplectic group into irreducibles and the tensor algebra you can tensor with itself four times and still decompose into irreducible so you can ask about what are the multiplicities of individual algebraic representations of the symplectic group inside this tensor product and now the striking fact is that the multiplicity of a given irreducible representation V lambda in this tensor product is given by polynomial in G of degree R times R plus 1 over 2 which is exactly the degree of this polynomial in the CFK or S conjecture which is not a coincidence and I should say something about how to interpret this fact that I just stated if I fix a partition of lambda I can think of that as indexing not just a single representation of some symplectic group but actually a consistent sequence of representations of all symplectic groups simultaneously so irreducible representations of this symplectic group I will index by partitions lambda 1 bigger than lambda 2 and so on bigger than lambda G bigger than 0 now if I have a partition of some finite length by padding it with zeros I can make it into a sequence of partitions of all possible lengths and that gives me a consistent sequence of irreducible representations of all symplectic groups simultaneously and once I fixed lambda any partition I can ask about its multiplicity when I decompose this guy into irreducibles and I let the genus G go to infinity and this quantity is the thing that is given by the degree r times r plus 1 over 2 polynomial and the polynomial depends on that and you see this using how duality but this is the key point but that's great I should also say by the way that if I have G small then V lambda doesn't really correspond to representation because here I said that irreducible representations of the symplectic group corresponded to partitions of this length and if lambda is of length bigger than G then V lambda is should not be thought of as an irreducible representation so V lambda is formally said to be 0 and then the formulas that I write down later are going to make sense but this fact that is given by polynomial of degree r times r plus 1 over 2 should probably be taken to be true when G is large enough that V lambda is 0 anyway that's all technicalities so to prove more logical stability for braid groups with these irreducible V lambda coefficients and now I think I'm in a position to state the main theorems of the two papers the theorem proven with Jeremy Easter let me write it down well the theorem with Jeremy Peter and Oscar is that this stabilization map is actually an isomorphism for K less than approximately n over 12 and I say less than approximately meaning that up to a constant K less than n over 12 minus 2 or something I don't remember and the point here is that this is independent of lambda so I told you that there's this theorem of Randall Williams and Wald that gives homological stability for families of discrete groups with polynomial coefficient systems and the stability range always depends on the degree of polynomiality and it's sharp for a general polynomial coefficient system you cannot get any bound that is independent of the degree and the point here is that if you specialize to this very specific situation where you have a family of irreducible representations of an arithmetic sequence of arithmetic groups and your coefficients are taken from the sequence of arithmetic groups then you can prove a stable range for this homological stability that has no dependence on the specific representation you choose and this is the key ingredient that allows us to control the error term in the main theorem and the second theorem that I want to say which is from this Bergstrom Diaconic Western paper is the explicit calculation of what is the stable homology with these particular coefficients and I'm going to write down the formula but it's not super enlightening I'm sorry to say it's given by a generating series so I sum over all partitions lambda all homological degrees k I take the dimension of the kth homology with a stable braid group on infinitely many strands with these b-lambda coefficients and then I take a Schur polynomial in infinitely many variables I take the transpose of this partition lambda so I flip the Ferrer's diagram around and I multiply this with a formal parameter minus z to the k so I track of the partition by using these Schur polynomials in infinitely many variables and I keep track of the homological degree using this parameter z and this expression now lives in I take the ring of symmetric polynomials in infinitely many variables I join an indeterminate z and then I have to take a completion because this is an infinite sum and the claim is that this formula is equal to something that I will write as follows x of the inverse log of 1 plus z sum of k bigger than 0 z to the k h2k minus 1 minus e2 and this x and log are certain operations called the pletistic exponential and the pletistic logarithm it's not the usual exponential and the usual logarithm which were introduced by Getzler and Kapranov in their work on modular operas and these are universal operations that act on complete lambda rings and this completed polynomial algebra over the ring of symmetric functions that's an example of a complete lambda ring so the takeaway is that there is this formula that looks a bit mysterious but it's actually quite computable you can give this formula to a computer and it will tell you exactly what are the stable homologies with coefficients in various v lambdas and it will do so quite fast so anything like Sage that can do calculations with symmetric functions will do this for you and you can say structural things about what a stable homology looks like and moreover the homology the braid group with these coefficients is pure k of weight I should write k here minus k plus the size of lambda minus 2k plus lambda but the Frobenius eigenvalues are q to the minus 2k plus size of lambda so I know exactly what all of the eigenvalues of Frobenius are on the stable homology and this is precisely what I need to estimate here I wanted to estimate oh sorry I wanted to estimate this as n goes to infinity and now I know both the dimensions of the stable groups and I know the Frobenius eigenvalues of the stable groups and one can give an infinite product expansion of this x log formula and this is you prove a general identity for the plastic exponential logarithm that gives you an infinite product which matches the Euler product from the CFKRS conjecture the Euler product that comes out of the recipe turns out to match exactly this x log formula that's proved by topological methods and once you have these two theorems it's not so much more work that has to be done to prove this theorem that I stated earlier so it's really these two give you the result the first controls the error term and the second tells you exactly what the polynomial is that it converges to my time is nearing and let me just say something about what goes into the proofs of the two so for the paper with Jeremy, Oscar and Peter we use this thing called cellular EK algebra and this is a recent invention of Sander Cooper's Seren Galatius and Oscar Randall Williams and it's a modern and multiplicative approach to proving stability theorems. I don't know if it would be possible to prove our result using this sorry, prove it without this cellular EK algebra but I do believe that they are essential and the key idea is that we want to prove this uniform stability for these particular coefficients that come from arithmetic groups and what we do is we lift uniform stability from the symplectic groups to the braid groups and the point is that we do know this uniformity for the stable range for the symplectic groups from Borel's work so Borel in the 70s proved exactly what is the stable density of sequences of classical arithmetic groups and in his work he proves this for coefficients in any irreducible representation and he gives a stable range that's independent of the specific representation which is exactly what we need for the braid groups and the goal then is to lift this uniformity from the symplectic groups to the braid groups and this can be done by writing down an appropriate EK algebra that measures the difference between these two groups in a sense and something about what goes on in the other paper is that we are inspired by what was proven by Madsen and Weiss proof of the Mumford conjecture Madsen and Weiss proved the Mumford conjecture which calculates the stable homology of the mapping class groups as opposed to the braid groups and there are now several different proofs of the Madsen Weiss one is via scanning and the scanning proof which is due to Galatius and Randall Williams is the one that we try to adapt to our setting and Madsen and Weiss were interested in constant coefficients but one can adapt what they did to polynomial coefficients and non-trivial coefficients and this was explained in a paper of Randall Williams and we try to adapt it to this hyperliptic family and it turns out that many of the same ideas work and in a sense they are simpler and some other things are more complicated and from this you can eventually derive this generating series and if you do things using logarithmic geometry you can make the constructions Galois be variant enough to compute the Galois action of the stable homology as well this was a very short proof sketch or it was not a proof sketch it was a list of ingredients but I'd be happy to say more about it I think I will stop here, thanks for listening


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Sonny Betty is here with me. He's the CIO and Chief Data Officer for Snowflake. Sonny, thanks for making the time today. Good to see you. Same here, Dave. Thanks for having me over. Yeah, so you're welcome. So before we get into it, I gotta ask you, I mean, you recently left NVIDIA to join Snowflake. I mean, one of the few companies that are almost as hot as Snowflake. How come? Well, you know, Dave, I joined NVIDIA 12 years ago. I was there for 12 years when NVIDIA was less than 2,000 people company and NVIDIA, you know, had an unbelievable growth trajectory. I went from 2,000 employees to 16,000 when I left in December of 2019 and Snowflake kind of provided the same opportunity to come in and help scale the company. I thrive in an environment where I can be creative. I thrive in an environment where I can build things. I can scale things. I can grow things and it's been just a perfect opportunity to come and repeat that success over here. Awesome. Well, we wish you the best. Talk about your role a little bit. I mean, it's not totally unique. I mean, especially in certain smaller organizations that have the same person in the role of Chief Information Officer and Chief Data Officer, but which are you? Are you more CIO, CDO? How do you balance that out? I would say that I'm both. To be an effective CIO, you need immersion with automation. You need immersion with data. You need immersion with security. And you also need immersion with compliance. So if all these things are together, things are integrated. You have a cohesive way of handling all the pieces that come together. We believe if you keep them separated, you create silos. And we definitely don't want silos. We want integration. We want seamless integration to drive and scale the company for our future. I always felt that my time is balanced between both areas. I mean, I always felt like a lot of the CIOs I talked to, they'd love to get more involved in the data, but they're just too busy trying to keep the lights on. So maybe what are your thoughts on the priorities of each hat, CIO and CDO? Yeah, so look, I mean, I think because we're a full cloud company, we don't have anything on-prem. I don't have any workloads on-prem. We don't have a data center. I really don't have to worry about all the operational challenges that you have to deal with being an on-prem company. So the cycles that I can be involved from a transformational perspective, driving transformation for the company, both on the data side as well as on the IT side, I have that cycles to invest that time and energy into both areas. Typically in a traditional company, which has not yet migrated towards the cloud, a major portion of their bandwidth gets wasted. CIOs bandwidth and IT professionals bandwidth gets wasted in dealing with the operational challenges that you have in an on-prem environment. So having not to worry about that over here gives me all the cycles to be investing my time in both areas. Yeah, a lot of wasted IT labor over the decades. Let me ask you, how is running a data company? We were inside of a fast-moving Silicon Valley tech company. What are the similarities and the differences from some of the customers? I mean, on the one hand, you're moving faster than your customers, at least most of them. And you don't have the technical day. You just described CXL Nirvana. On the other hand, you're an example of what's possible. You can sort of set the best practice mark. How do you see that dynamic? So for a world-class IT organization, it needs to be data-driven. It needs to be highly automated. It needs to enable world-class user experience and then to secure and make the environment compliant and resilient. The cloud platform that we have inside Snowflake allows us to achieve all of that. Now that is an ideal situation to be in where you don't have to deal with all the on-prem type of workloads. So finding the balance is what we're going after. However, this is a journey. For other companies who are not on the cloud, it's a journey. They have to prioritize that. They have to start moving things to the cloud. And that's where we are different and similar. We're different that we don't have to worry about that. Everything is in the cloud for us. And then that's how we see it. So I used to call it the dog-fooding segment, but Oliver Busman was the CIO of SAP. He said, no, Dave, we call it drinking your own champagne, which is how you guys refer to it. But sometimes, still in such situations, you're inside the sausage factory, which is good in a way because you see it before it goes into production. So what's your journey with Snowflake been like? Yeah, so that's a really good question. That's a major portion of what I do at work. And let's start with the first principles. We believe that we want to measure everything in the company that's important for companies' performance. If we measure the right things, we believe we can drive the best outcomes. We are driven through those first principles and we leverage our business applications, our data, our security, our automation, and our compliance to integrate with our product to power all these use cases and workloads. In our own environment, we call that Snowhouse, which is nothing but a Snowflake instance. So for all the new products that we are coming into market with, we work very closely with the engineering team, with the product management team to make sure that we actually become customer zero and try to use as much functionality of that inside our own enterprise and give as much feedback to our engineering and our product management team so that they can make the customer one experience to be world class. So that's kind of in a nutshell how we go to market with those products.  I think it's a very mutual beneficial relationship. So they know that my team's feedback is important to how they're kind of shaping up the product. And it's just not necessarily IT, right? We have folks in finance, folks in sales, marketing. Everybody is drinking the champagne, right? And IT and the data team actually enable that deployment, but the use cases are pretty much in the entire enterprise of the company in every aspect of it. Well, you know, including security. Well, you know, as I was just saying, we always talk about alignment, but it's like it's almost alignment by design as opposed to being this forced thing. I'm interested in this sort of snowflake on snowflake concept that you guys talk about. What were your objectives going in and maybe thinking about the outcomes? What did you expect? Did you work backwards from that? What were you trying to achieve? Yeah, I mean, look, again, back to the first principles. We believe we want to measure everything that's important to our business that would drive the right outcomes. We then layer the application layer. We then overlay the business process layer. We then overlay the compliance and security layer. And the end result really is operationalizing snowflake internally to drive our business, making the right choices, right decisions for the company. So we have a ton of use cases that are just ideal using snowflake on snowflake. And I can give you some examples of that if you like, but security being one of the biggest use cases, we use the entire monitoring and remediation work that goes in the security compliance world all through snowflake. And we are finding real-time events through data sharing with our key suppliers. And we're ensuring that we're protecting our environment as much as possible with that whole infrastructure. We talk about layering, governance, security, et cetera. I'm imagining a coat of primer paint, nice and smooth. It's not a bolt-on.  And what you're describing is much more of a modern approach. And I want you to differentiate between the layers that you talked about and what you've surely seen in your experience over the years as a bolt-on. What's the difference? Well, I mean, the security world, there's a lot of data. And a lot of the data that is critical to your environment, you want to make sure, A, it's fully complete. You're getting it in the right hands, in the right platform to understand and doing the correlation work that needs to happen real-time. Our platform allows all that data to be ingested in real-time. And anything that is suspicious that's being out there, we're finding that stuff in real-time, the monitoring has to be real-time. And if there is an event, somebody needs to take an action real-time. So the platform allows it to integrate all together. And basically, the suppliers that we're using are also doing data sharing with us on this platform. So it makes the whole security remediation to be really, really fantastic experience. Well, I think too, I share often with my audience, when I talk to practitioners that are using Snowflake, surprising to me when I first heard this, they said, well, would you show Snowflake because of the security? I went, what? But the simplicity and the workflow is simpler. And it just means less human labor involved in setting these things up. So I wonder if you could talk about the team that you put together, the culture that you're building, and what's the makeup look like? Sure. So are you specifically asking about the characteristics of how we're building up the culture? Yeah, absolutely. Okay. So I think we're looking for obviously very much high-energy folks, people who have high accountability, they're data-driven. We want to measure everything that's important to us. We're looking for folks who have situational awareness. And then finally, high sense of urgency. I think all of these elements allows IT organization to be integrated with the business. In a lot of the traditional companies, IT organizations kind of disintegrate with the business. We want to integrate with the business to drive the best outcomes that are needed for the company. I want to ask you about some of your favorite use cases. You mentioned measurement. How do you measure? What are you measuring? Sure. So I would say that let's just take security because we talked about security. Let's just use security as a use case. So in security, there are many different frameworks, as you may know. There is the Nest framework. There is the CIS framework. There is an ISO framework. We have adopted towards a CIS framework inside Snowflake. That framework has 20 controls, and that 20 controls has another 20 sub-controls. So we're talking about 400 controls potentially. Not every control is applicable to us, but the majority of them are. And so for every control, there is a source of data that's being ingested in Snowflake. I'll give you an example of that is asset management. So asset management for endpoints, asset management for our servers, or asset management for our network gear, all of that data gets ingested inside Snowflake. We measure that. We can tell you exactly how many endpoints I have. I can tell you exactly when an employee gets onboarded. What laptop we have given them. When the employee leaves the company, are we collecting that laptop back on time? Are we working all that access? That's part of CIS control 1 as an example. And we're measuring all of that. And I can tell you exactly at my real time inside Snowflake how effective I am for that specific control. That's just an example of that, Dave. Now imagine 400 of these items that make up the whole security CIS framework. You want to measure everything on that 400 controls or 400 sub-controls. And you want to make sure that if any of that control is not being managed properly, you're alerted about it, and you're remediating it to prevent a security issue that may pop up. Awesome. Visibility and automation component. Are you the CISO too, Sonny? We don't really have that title. We don't really have a CISO title, but I do where the security had as well. It's actually a joint responsibility between... I manage the corporate security. The product security is inside the product team, but we use the same common framework. We use the same common telemetry. We use the same common methodology. Incident management response teams are very similar. And it's all powered through Snowflake. Okay. And thank you for watching. Keep it right there. We've got more great content coming your way.


Hi beautiful You look whoa you Come here. Come close Come close. Come here. You're beautiful. I mean that. Yes, you don't do that. Thank you for for joining me today I really appreciate it. We're watching some butterfly cuts and we're gonna fly away like a butterfly together I'm sure a lot of people are gonna do them completely wrong because that is always what happens You know the tea but first I want to thank care of for collaborating with me on this video You know, I love care of they've been in so many videos And we've had such great partnership over the years and you guys also really enjoy their products And if you don't already know what care of is well, I'm here to explain it to you Care of is a subscription service that ships high quality personalized vitamins supplements and powders to your door every month And if you're like well, I don't know what supplements and vitamins I need care of quiz takes all the guesswork out of building your routine The quiz asks you general questions about your diet lifestyle and health goals Then it recommends the right vitamins and supplements for your specific needs and goals a big goal of mine this year is to get So damn physically fit like look divine like a little snack noodle Obviously, I needed the right supplements in order to do that to keep myself healthy and energize while going through that process So my cute little pack here obviously has vitamins in it that will help with that. It'll make me strong It'll make me energize. Let me feel good throughout the day I do Pilates and listen you need a lot of energy for that Oh, it takes a lot of mental strength to get through an hour long class four or five times a week in my pack There is be complex which helps with energy and magnesium for bone and muscle health And these packs are made with plant-based film, which makes them compostable I personally trust care of because they use clean ingredients that are backed by science They deliver the vitamins right to my house and I don't have to buy a bunch of different packs of vitamins and then make my own Concoction with it. It's all like already in this compostable pack which makes everything really easy for me and they have a new product It's called gut musk This is their latest line of versatile formulas that level up your existing routine to manage bloating gas and digestion Weight take care of quiz and see what vitamins and supplements they recommend for you Click on the link below and use my code Brad Mondo for a 50% off your first order with care of thank you Up first we have anybody Alicia Monak. Here's my laptop with the Brad Mondo birthlight cut tutorial We are following my tutorial we're going in there and it's gonna be great. She's gonna do it I'm trying to give myself a face framing layers I don't like how flat my hair is because it's already long Oh, I think she would look so good with a butterfly cut her hair is full. She's a good amount of hair You know, it's not like it's not like whoa. It's not like pounds and pounds and jebel. It's like pounds of hair It's like a good one pounder of hair. I'm trying to tame myself today. Um, wow. What is this a magnet? No, it's battery. Hmm a little more Of this and then maybe that sentence will come out. Ah, yes, and if she adds some layering in there It's gonna like pop and lock and what else? Drop and it's gonna look great. How much I cut it. I'm really scared She got the Ariana Grande pony in the back Which is always what we want and she pulled all the hair to the front without a bunch of bumps and lumps and her hair is very Pin straight. I would have recommended. I'm pretty sure I always recommend in my videos I hope so Brad. Do you you recommend everybody wet their hair before they do these cuts, right? So she didn't follow that direction F I have no idea what I'm doing How much I'm supposed to cut the thing is that people often don't cut a lot And then they don't see any results and that's not what I want for you Like I honestly I would love for you to cut a lot and get good results I don't even know what kind of haircut I want, but she's using hair cutting scissors. Yeah a plus plus plus I'm doing. Oh my god. I'm doing That's actually so much not enough. It looks like you got air cutting scissors, but you're supposed to get hair cutting scissors I was a little funny. Come on working Because that's only cutting air not hair. Okay. I dig it. It's not gonna look bad, right? Am I supposed to do this this little thing right here?  God I don't know if I'm doing this right. I'm gonna cut just this cuz I'm scared girl It's gonna be like one of those really boring ones with is don't cut anything off because I'm just not interested I don't want to watch it if it's not a lot of hair being cut. How much I cut it? It was a half an inch. I'm regretting my decisions. We also got to cut the same amount off the back as we do on the front. Okay.  I need to see something happen today. Come on I think I was supposed to cut a little bit more. Yes So what are we gonna do about it? Are we gonna cut more? Are we gonna like check it out? Are we gonna do actually do the haircut because this this is gonna make any difference? Okay? There's a lot of hair. I need to brush my hair first Do more you should always cut off at least four inches or else. There's gonna be nothing to look at I want to cut a little bit more. I think this time. I'm gonna cut a little bit shorter like this Yes, see now we're doing something now you have my interest Those scissors hurt my ears.  It's a lot of hair Remember you guys you are cutting layers.  This is the part where we all want to always have is the lengthy poo Should I cut a little bit more? Oh, yeah, keep going. I think this amount of hair It's enough as I say, I don't want to lose the length of my hair I just want to have a little different cut. Okay now. She's good.  Oh my god so much hair. I think it's enough. Let's see the results Let's check it. It's always good to check what you've done and then go back and do more. Oh It looks bad. I need to style first to see what I just did, but it looks very It looks great.  You have to like then wash it This is why you recommend cutting in a wet hair because it's just like makes more sense that way It's not gonna look appropriate right now. I'm gonna try to Cut to make a little bit softer use some thinning shears This would really speed up the process instead of like trying to point cut it with like dull air scissors Take out more hair at a time and it'll make it so much softer. Okay to see the better the results I'm gonna style my hair with a flat iron That was a huge glow up. That was the best Decision of my life. Look at my hair with the layers. Oh my god. It looks so pretty Did she got a beautiful haircut? It looks so good. Thank God. This haircut is like foolproof mostly. It feels good. Oh, oh Oh, I did it this straight. You slayed the day away Congratulations, you win a great job and next time.  We're about to do something very drastic. I want to try this wolf cut I'm already have I got the wolf look going right now in my roots. I grown out. Yes You do got the wolf look going right now. I love it though.  So that's why it looks extra Moxie up here. She could definitely definitely benefit from some layering. It is heavy You know, you see how it's like boxy it goes like this like this and like this That's because there's nothing in between to round out that hair and she needs that but I want to try this Oh my god, I can't wait to see those dead ends just drop to the ground to the floor and just lay there Like a little dead end It's gonna be so good for her like look at all those ends that are just like I wish I was there to cut this I don't know why it just like I can see that like puff ball of hair I just want to like like how satisfying of that beat She went from ear to ear all the way around the back and she pulled that hair nice and talk forward with no bumps Perfect, but she's failed to wet her hair before she does this cut Ah, I suppose it would be better for her to cut it on straight hair if she wears her hair mostly straight rather than curly Yeah, I'm about to really chop my hair off. That's half of the ponytail Oh, did you see those ends? They were like That's what I imagine they would be saying if they could speak and she's about to go Let's do it Freakin do it girl. Yeah, except not enough hair add more hair and cut more It's like on one hand. I want you guys to start off small, you know work your way up on the other hand I just want to see a fun result really quickly Don't point cut it just cut it straight across you're gonna get like lines going like this If you don't point cut correctly and that's what it looks like it's going on and also I couldn't even see the cut the best part was blocked by the view of the camera All right at that point just cut straight across it's like So that is a no go Don't tell me you're gonna leave that one little hair out like that. What is that hair? Everybody's point cutting so Vertically into the hair that it's not even taking any hair out You know like if you're gonna point cut and you want it to make a difference You got to have a little bit of angle to your scissors to take out a little bit of that hair in between If you like this you're gonna take out one strand at a time and only like hair cutting pros know how to do that quick enough To actually make a difference. It seems a little blunt still We actually did take the hair a little bit too far in the back towards the front Like we should section the hair from the apex of the head This is the tallest point of the head her hair was taken from like all the way back here forward And we don't want that because we want a lot of layering in the back So you get all that volume if you take the crown hair forward It's gonna give you the least amount of volume and texture at that crown area So make sure you don't take it that far back She's talking about me he said pull it to the front I did And also your fingers are again like this straight we got to go straight So stressful I was gonna say is she not seeing this but obviously not she's not seeing the fact that it is completely Diagonal the line. She's cutting. All right, so I know the hair isn't gonna look completely Awesome tonight, but I feel like something has happened I'm just gonna brush it down Track to feather it back so you guys can't see okay Definitely dry it's really freaking fine. Okay. It's definitely pretty. She's a lot more movement But like she's got to do more and she she's going to there's still time left in this video I'm gonna just like do halfway down the pony tail That much that looks so much better this time the sectioning is so much better. Yeah, we're still cutting the hair Completely crooked. This is definitely a first. I've never seen anybody cut something quite this crooked before It's like she's pulling the hair up and then she goes like this next time just Cut it straight. I mean Okay, I'm gonna blow myself up get some type of a shape going on so y'all can actually see it Hey guys, so this is what it's looking like, you know, every look comes together. We're to be games in a day It's not enough. I think I got a lot more here I feel like I might have cut a little bit more But then when I'm looking at this side this side looks better when it's like straight, but I'm loving those results Look so good on her her hair looks so much more full of life body movement. It's so much better Oh, she looks amazing and with the curls with the with the layering with the texture and her personality and her makeup It all came together really really really good. Congrats. You look amazing up next. We have a video by purple potion I am so tired of my hair. Okay, it looks cool. I want layers. Oh It's just alright like she needs layers layers are the cure to sadness and depression Yeah, I'm just gonna like become a therapist and just prescribe layers to people because it really does help with everything I saw this butterfly cut a falling bread model He said to put the comb on top of your head and find a high point aka apex So I just take it and it's going to go down to behind the ears Her hair is wet. She followed the tutorial with the apex or onto something good I'll come to me if it's not perfect because I use glasses normally. Oh great. So we're gonna cut blind Basically, awesome. The thing is he takes the ponytail hair, but I have some experience cutting hair I'm a wig maker professional.  Teach me. I love wigs. He said okay, sweet here I know better. I do just drag and take one line of designer cut then put The other one the fold line for sure. You can totally do it like that I honestly think just putting in a pointele gives you pretty much the same exact effect and it's much easier You can take the hard path too if you want. I have a bang That is almost grown out. I would cut a little bit. I would just drag it straight out like this We chose the point cut again I think the point cut is like the safe space for people, right? It's like I he's cutting shirt across feels like a lot more than doing a little pointing So but let's retire the point cut for for at-home tutorials. It never comes out great like that You know, she's definitely doing a good technique and it's solid what she's doing But just let's retire the point cut. I wanted to do not be perfect wanted to be more Rockers but if I could so she made a guide and now she's following the guide for the entire front of her head just Do it She's getting that layering in the front that rounded look that we all love to see we're getting through it I wish you just putting a point on this cut it, but it's okay. I can't control everything That's cool, man. Haha. That is cool, man. We're taking more Yeah, this is Next step pull everything and go and that's way too much hair more and but okay. I'm doing it Look at this dude. Oh, really cool man. She's went from like cool to really cool very quickly Look at this hair. Yo, I cut a lot of hair off and it looks like I have even more hair But it feels lighter. I have not styled it actually I just put some volumizing powder at the root and my hair just laid like this Naturally, she was talking about how now her hair kind of just lays in the place that it's supposed to and if you do a good Cut on yourself or you go get a good cut your hair should lay in the perfect spot without you doing too much That's the finish look and I'm rocking it I think oh she looks so great the haircut with the color the look I love when you guys pull it all together and just make like an entire moment out of your haircut or your hair color It makes me so excited the important part is that I feel good. I don't care about what else That's exactly true and only matters how you feel about yourself, um, but I love it also that means anything I love to see all the butterfly cuts being slayed around the world It is so exciting and great job you guys if you want to check out my hair Carolina my hair color line You can do so with the links right down below or go to expondo hair comm that is all for today Thank you for watching. Don't forget to live your extra life, and I'll see you next time. Bye


NATO chief. Ukraine needs to be armed for a long war. Ukraine's foreign backers should be prepared to maintain their military support for a long time. NATO Secretary General Jens Stoltenberg has said. The need for weapons will continue to be there because this is a war of attrition. This is about industrial capacity to sustain the support. The official told the Guardian newspaper. He noted that Ukraine's expenditure of munitions in the conflict with Russia has outpaced the production capacity of the nation's arming Kiev. The NATO chief urged members of the US-led bloc to beef up their military spending to at least the target level of 2% GDP to sustain the war effort. Moscow, Stoltenberg claimed, is also running low on weapons despite boosting its manufacturing and is seeking arms from other nations including Iran and North Korea. Such claims have repeatedly been made by Western officials, also with reference to China. But Russia and those countries have all denied any such supplies. Stoltenberg also criticised Beijing's attempt to mediate in the conflict and broker a peace agreement. Beijing, he insisted, needs to understand Ukraine's perspectives and engage with President Volodymyr Zelensky directly. China-Ukraine relations reportedly suffered a serious blow after the Zelensky government blocked the purchase of Ukrainian aerospace giant Motor Seek by a Chinese firm. The deal was effectively stopped in January 2021 when Kiev imposed sanctions on the Chinese businessmen behind them prohibiting the transfer of assets. Earlier this week, German Chancellor Olaf Scholz issued a similar warning on the Russia-Ukraine conflict, saying Western powers should prepare ourselves that it can last a long time.


One previous Premier League campaign, they've finished rock bottom. Here comes the Cherries, they will send in some space in the penalty area now. Driven wide, pulls it back, Frazer! The best player in pre-season is the man who opens their scoring account in the Premier League. Ryan Frazer, 23rd minute, nice build-up down the right-hand side. As it was pulled back, the We-Man made no mistake, stroked it beyond Etherage. It's formed with one card at nil. Cherries working hard in centre of the field to win it back. You haven't missed anything by the way, that was a stoppage. The Cherries coming forward, Callum Wilson. In behind all the way, Wilson in the penalty area, topples down, it's a penalty! It wasn't a huge amount of protest, I have to say. No, it wasn't. Not clear cut, anyway. Kevin Friend gave it, Callum Wilson is going to take it. Waited a long time for that goal on the final day of last season to break his long droughts. Can he kick off this season on day one by hitting the back of the net from the penalty spot? It's Wilson against Everidge in the Cardiff goal, saved by the goalkeeper. Palmed away, down to his left. Ball up towards the far side, Cherries still going. Francis battling away, right side of the area, into the box goes Francis. Still going to the ball, like, Wilson! Yes! That missed opportunity earlier on, and the Cherries put it to bed. They spent a sticky 20 minutes or so with a bit of threat from Cardiff, but Bournemouth have been the better side overall and will kick off the season with three points on the opening game. Bournemouth to Cardiff nil, Callum Wilson.