06.06.23 (Koszul duality for operads and the logarithm)
Hi everyone, long time no hear. Currently I am busy preparing all kinds of talks and life keeps on going, however I want to quickly write down some really nice thing (that can for instance be found in Loday-Vallete Algebraic Operads) - namely that Koszul duality for operad means invertiblity composition-wise. Take for instance the associative operad, the associated power series comes from the free associated algebra and is the 'power series version' of the free associative (non-unital) algebra (The tensor algebra!). The power series is just x+x^2+... = x/(1-x). That the associative operad is Koszul self dual implies that if we take this power series, replace x by -x, put an additional (-1) in front (i.e. here we get x/1+x) and insert it into the original power series x/1-x we get the composition neutral element x. Similary the Koszul duality of the commutative operad and the Lie operad implies a similar statement. The commutative operad gives the power series exp(x)-1 = x+x^2+x^3/6+... and it's composition inverse is log(1+x)=x-x/2+x/3-x/4+..., i.e. the Lie operad has corresponding power series given by -log(1-x)=1+x+x^2/2+x^3/3+.... Now from this calculation alone you see since the coefficient in the power series corresponding to an operad is the dimension of the n-ary operations corresponding to that operad divided by n!, you see that Lie(n) has dimension (n-1)! since 1/n = (n-1)!/n!. One should think of this is an analogue of the Koszul duality invertibility criterion for graded Cartan matrices that can be for instance found in 'Koszul duality patterns in Representation Theory' by Beilinson, Ginzburg and Soergel. Enjoy the rest of the spring, before the eventual heat comes to destroy us. This is also a NBA finals reference - Go Nuggets!
13.04.23 (Grassmannians vs vector bundles)
The more I write stuff in this blog the more I feel like a geometer/geometric representation theorist. Today I learnt something in Andreas Thoms K-theory lecture, which blew my mind. So we proved that every vector bundle (we are in the typical real geometry setup) on a compact space is a direct summand of a trivial bundle (Think: Every projective module is a direct sum of a free module). The rank of your free module/trivial bundle can be chosen as the numbers of opens you need to trivialize your bundle times the dimension of the bundle. However the madness starts in the next paragraph (which every topologist probably knows but suprised me).
So basically the mindblowing statement that I learned was that Grassmanians control vector bundles at least on compact spaces. So when you take the Grassmanian Gr(m,n), whose points are the m-dimensional subspaces of an n-dimensional space, it has a wonderful bundle -> the tautological bundle. Now when you have a morphism f: X->Y of spaces and you have a vector bundle E->Y you can take the pull-back to construct a bundle on X. This might not sound so crazy yet, but now notice that for any space X together with a m-dimensional vector bundle E->X, which is a direct summand of the n-dimensional trivial bundle, we can construct a map X->Gr(m,n). How do we do this, you might ask. Well take any point x in X and map it to it's fiber in E, which is an m-dimensional vector space, which we can view as m-dimensional subspace of k^n (You can do this since by assumption E is a direct summand of the trivial bundle). Voila: This construction shows that m-dimensional direct summands of the n-dimensional trivial bundle on X correspond to maps X->Gr(m,n). Now you might ask: What if I pull-back the tautological bundle on Gr(m,n) back to X: Well it becomes the bundle you started with!!! Crazy stuff, but normal in the daily life of topologists. And not just topologists, but algebraic geometers also as Timm told me. If you don't know how to define the Grassmanian over any ring, just take the functor sending any scheme X to the set of all of all surjective sheaf homomorphisms O_X^n->V, where V is some locally free sheaf of rank k. The object representing this functor will be the thing you want to call Grassmanian Gr(k,n).
Anyway what should we learn from this for our future lives? It tells us that vector bundles are in some sense a Dynkin type A notion. In particular it is clear that at least on a compact space X a vector bundle should be a morphism X -> G/P where P is some cominiscule parabolic.
12.04.23 (Min-Matrices, Highest-Weight-Stuff and Koszul algebras)
So here is a beautiful thing I cooked up a few months ago, maybe I can someday write a paper about it. Let me start with a story. I have a guy living with me in my shared appartment, let's just call him Max or rather Min. It will make sense in a moment. So I told him I do a math phd and he told me he is also doing a phd, but in electro engineering. But then he actually explained to me what he was doing and it turned out to be a lot of linear algebra, basically trying to compute certain diagonal entries of some crazy matrices.
Basically the goal was to take a min(i,j) matrix (i.e. the matrix whose i-j entry is min(i,j)), add a scalar multiple of the identity, then take the inverse matrix of the result and then add to that inverse an identity matrix again and compute the inverse. What would be enough: What are the diagonal entries of this matrix? What happens when the matrix gets arbitrary large? This is some crazy problems to tackle. If you want an easier, but still research worthy problem: Just take the min-matrix, add a scalar multiple of the identity, and invert the result. What will you get explicitely?
Anyway that's not what I will talk about. However there are three crazy things to notice with min-matrices :
The determinant of them is always 1
They factor as a certain nice product of an upper triangular matrix with 1's everywhere on the diagonal and on top of it and it's inverse
The inverse has a nice description with many zeros (Explicitely it is pretty much the type A_n Cartan matrix, but you replace the last (i.e. n-n-th) diagonal entry by 1 instead of 2.
So when a woman/man of culture reads this they smell Cartan matrices: Determinant 1 - aha, the corresponding finite dimensional algebra has finite global dimension. They factor as some nice product of an upper triangular matrix and it's transpose --> BGG reciprocity! The inverse has a nice description - Koszul duality!
Anyways I will not spoil the finite dimensional algebra behind it, you can try to find out for your own! Hint: The min matrix for n=2 is the algebra describing the principal block of sl2...
09.04.23 (Some observations on Dynkin folding)
Hello, so there is all this talk of folding of Dynkin diagrams whatsoever. Of course one knows how to do this in the Lie algebra case: If two vertices of the Dynkin diagram get folded into one, take the sum of their Chevalley generators. For the algebraic this should mean take the product of the Chevalley generators. But what does all of this mean to a down to earth person just understanding basics on Coxeter groups?
Well whenever you fold a Dynkin diagram into another one, take the products of the Coxeter generators you folded onto one another and this should be the Weyl group of the folded result. Warning: For this to work you are only allowed to fold Coxeter generators that are not neighboured, i.e. the corresponding simple reflections commute. This is just as for Lie algebras and algebraic groups.
Anyways let's give some cool explicit examples of folding:
A_{2n-1} --> B_n: This realizes the Weyl group of type B_n as sign symmetric permutations, hurray!
B_3 --> G_2 This one is really nice: One realizes G_2 as subgroup of B_3 generated by s_0s_2 and s_1.
D_n ---> B_{n-1} In terms of dotted permutations this is a cool one: Take all dotted permutations on n strands with an even number of dots (This describes D_n) and embedd B_{n-1} by shifting all simple transposition generators s_1, ..., s_{n-1} by one to the right and identify s_0 with two dots on the very left strand
Something that one observes in all these examples: The longest element in the subgroups is actually the longest element of the big group, but of course the length changes since the embeddings are not parabolic. An interesting question would be: First, how does one show conceptionally that all of those are embeddings instead of doing case by case arguments. Second: Why should the longest element be mapped to the longest element? Similarly does it define an embedding on the braid group level? If so this should imply that the full twists are mapped to one another under folding, which should imply that in some appropiate setup the Jones--Wenzl projectors are the same. Lots of questions.
08.04.23 (Eastern special: Special orthogonal groups over finite fields, the type B Weyl group and the field with one element, part II - Cardinalities of spheres)
This is a continuation from downstairs. If you are a topologist/algebraic geometer you might like this section. So recall the cardinality of the type B_n Weyl group (as always the symmetry group of the hypercube imagined as generated by n reflections turning it into a Coxeter group), which is 2^n*n!. Concretely it is generated by n elements s_0, s_1, ..., s_{n-1} which satisfy the relations s_i^2=1, s_is_j=s_js_i if |i-j|>1, s_0s_1s_0s_1=s_1s_0s_1s_0 and s_i*s_(i+1)*s_i=s_(i+1)*s_i*s_(i+1) for i>0. The type D_n Coxeter group (of cardinality 2^(n-1)n!) can be regarded as an index 2-subgroup of the type B_n Weyl group, it has generators s_0', s_1, ..., s_{n-1}, where s_0'=s_0s_1s_0 under the mentioned inclusion (which leads to s_0' and s_1 commuting and s_0' and s_2 satisfying a Reidemeister III relation). I will try to convince you that the index 2 is the cardinality of an even sphere over F_1. Similarly the cardinality of the type D_n Weyl group being 2^(n-1)n!, which is n times more then the cardinality of the B_{n-1} Weyl group (being 2^(n-1)(n-1)!) is no coincidence.
In order for the statement to make sense I work with a finite field F_q of characterstic not equal to 2. [WARNING: There is an important subtelty that I need in order for my formulas to work, which is that a square root of -1 doesn't exist in my finite field F_q. So for instance F_3 doesn't have such a root, F_9=F_3[i] has one, F_27 doesn't have such a root and F_81 again has such a root. One can show that any finite field, the existence of i is equivalent to (q-1) being divisible by 4, since the group of units in F_q is isomorphic to a cyclic group. The subtelty is that technically there are two non-isomorphic SO_2n(F_q) depending on the non-degenerate symmetric bilinear-form one chooses of F_q^(2n). One special orthogonal group corresponds to the identity matrix and is denoted SO_(2n)^+ and one corresponds to a size (n-2) times (n-2) identity matrix with a (2x2) minus identity matrix attached denoted SO_(2n)^-.]
Now recall from your generic math knowledge that we have a fibre bundle SO_(m-1)-> SO_m->S^(m-1) over the finite field F_q. Which implies that the cardinality of S^(m-1) is given by the quotient of the cardinalities of SO_(m) and SO(m-1). First let's look at the case where m=2n+1 is odd. In this case we get the fibre bundle SO_(2n)-> SO_(2n+1)->S^(2n-1), so in the middle we have type B, on the very left type D and on the right a nice sphere. Let's now compare cardinalities using our calculations from last time:
|SO_m|=|SO_(2n+1)|=q^(n^2)*(q-1)^n*(1+q)(1+q^2)...(1+q^n)[n]'_q[n-1]'_q...[1]'_q
|SO_(m-1)|=|SO_(2n)^+|=q^(n^2-n)*(q-1)^n*(1+q)(1+q^2)...(1+q^(n-1))[n]'_q[n-1]'_q...[1]'_q
So we get |S^(2n)|=|SO_(2n+1)| / |SO_(2n)| = q^n (1 + q^n)=q^n + q^2n, setting q=1 gives the desired 2.
Now let's look at the other case, where m=2n is even. Again let's compare cardinalities:
|SO_m|=|SO_(2n)^+|=q^(n^2-n)*(q-1)^n*(1+q)(1+q^2)...(1+q^(n-1))[n]'_q[n-1]'_q...[1]'_q
|SO_m|=|SO_(2n-1)|=q^((n-1)^2)*(q-1)^(n-1)*(1+q)(1+q^2)...(1+q^(n-1))[n-1]'_q...[1]'_q
So we get |S^(2n-1)|=|SO_(2n)| / |SO_(2n-1)| = q^(n-1)*[n]'_q, voila!
Final remark: All of these calculationshave to be adjusted if a square root of -1 exists in F_q. Then for instance one can show that |S^1(F_q)|=q-1 instead of |S^1(F_q)|=q+1, which we calculated. If you want to figure this out yourself: Use the factorization of x^2+y^2=1 as (x+i*y)(x-i*y)=1. The question to you: How many points should S^1 have over F_1? Two (=q+1, square root of -1 doesn't exist) or rather zero (=q-1, square root of -1 exists)?
08.04.23 (Eastern special: Special orthogonal groups over finite fields, the type B Weyl group and the field with one element, part I - Cardinality of SO_(2n+1))
Happy Eastern! Remember how I mentioned on the 31.03.23 the rank function giving the length of stuff in the type B_n Weyl group (fancy word for symmetry group of a n-dimensional hypercube) - like (1+q)(1+q^2)...(1+q^n)[n]'_q[n-1]'_q...[1]'_q, where [k]'_q=1+q+...+q^(k-1). These sort of functions are cool, since if you insert q=1 they give the cardinality of the group, but even further they tell you how many group elements of a specific length (say length k) you have by reading of the coefficient in front of q^k. For type B the formula tells us that the highest exponent of q is n^2, which is precisely the length of the(!) longest element. Well I asked myself how this is related to the cardinality of the type B semisimple connected algebraic group SO_(2n+1)(F_q).
In order to answer this question let's look at type A, which features our favourite objects: GL_n, SL_n, upper triangular matrices, symmetric groups, flags of vector subspaces. One way to get to the length rank generating function for type A_{n-1}, i.e. our favourite group the symmetric group S_n, is by counting the number of elements in the full flag variety over the finite field F_q, i.e. the set of all vector subspace flags in F_q^n. The number is given by [n]'_q[n-1]'_q...[1]'_q,, where you have [n]'_q possibilities to choose a 1-dimensional subspace of F_q^n, then [n-1]'_q possibilities to turn it into a 2-dimensional subspace, then [n-2]'_q to turn this into a 3-dimensional subspace and so on until there is only one way to increase it to an n-dimensional subspace. The intuition for this is that S_n can be identified with the set of subset flags of n elements: You have n possibilites for a cardinality 1 subset, then (n-1) possibilites to choose one more element turning it into a size 2 subset, etc. etc. until there is only 1 possibility left. Think of flags of sets as vector space flags over an imaginary field F_1 with 1 element.
Well known thing: You can identify the full flag variety with the set GL_n/B, where B denotes the set of upper triangular matrices. This identification shines new light on the cardinality of the full flag variety over a finite field, since the cardinality of GL_n over F_q is given by (q^n-1)*(q^n-q)*...*(q^n-q^(n-1)), which can and should be written as q^(n choose 2)*(q-1)^n*([n]'_q[n-1]'_q...[1]'_q) and the number of the upper triangular matrices is q^(n choose 2)*(q-1)^n, so by GL_n by B we get [n]'_q[n-1]'_q...[1]'_q. Voila! We get our rank generating function of the symmetric group. What is behind this, if you wanna look it up: Bruhat decomposition.
Anyways let's observe some things in the formula for the cardinality of GL_n and the Borel B and try to guess a formula for the cardinality of SO_{2n+1}(F_q), i.e. the type B connected semisimple algebraic group of type B over F_q:
There is a q^(n choose 2), which is precisely q^(size of all strictly upper triangular matrices), which is precisely q^(length of the longest element in S_n). If we do this for SO_{2n+1}(F_q), we expect q^(n^2), since n^2 is exactly the length of the longest element in type B, which is the number of positive roots in the type B root system.
There is a (q-1)^n, which counts the number of diagonal matrices (also called the torus). If we replace GL_n by SL_n, we get here (q-1)^(n-1), which makes sense since SL_n is Dynkin type A_(n-1). So we expect (q-1)^(number labelling the Dynkin type)=(q-1)^(rank of the torus)=(q-1)^(number of simple roots), etc. etc. Anyway following reasoning we expect for type B_n a factor of (q-1)^n in the cardinality of SO_{2n+1}(F_q).
Finally there is the length rank generating function for the type B Weyl group, which we also expect, so a factor of (1+q)(1+q^2)...(1+q^n)[n]'_q[n-1]'_q...[1]'_q.
Concluding this we expect that the cardinality of SO_{2n+1}(F_q) is the product of all these three ingredients, which is q^(n^2)*(q-1)^n*(1+q)(1+q^2)...(1+q^n)[n]'_q[n-1]'_q...[1]'_q, which is exactly the formula from wikipedia rewritten: https://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields (note that they write about the orthogonal group so there is a factor of 2, since SO_n is an index 2 subgroup of O_n, unless the characteristic of the field is 2, in which case SO_n=O_n). Since this diary entry got so long already: I will continue in a second part.
05.04.23 (Continuation)
Apparently one always has if A is Hopf algebra and M is an A-bimodule that Ext_H^e(A,M) is isomorphic to Ext_H(k, adj(M)). The top-side lives in the Hochschild world and the bottom in the group cohomology/Lie algebra cohomology alikes world. Here adj(M) is the H-left module obtained by turning the left-right action of H on M into a left action via precomposing with the algebra homomorphism H -> H otimes H^(op) given by x \mapsto x_(1) \otimes S(x_(2)). In particular the group cohomology of kG (with the adjoint action) & Lie algebra cohomology of U(g) (with the adjoint action) are precisely the Hochschild cohomologies of kG respectively U(g). Applying the Lie algebra case to g=k^n the n-dimensional abelian Lie algebra, gives the easiest case of the Hochschild-Kostant-Rosenberg theorem: HH(k[x_1,...,x_n]] is given by exactly taking k[x1,...,x_n] many copies of the ext-algebra of the trivial module with itself, which is precisely the exterior algebra in n variables (which is here the Koszul dual).
05.04.23 (Thoughts on center of Hopf algebras, universal enveloping algebras and Representation Theory as a whole)
Connected to the last diary entry: For a general Hopf algebra H one can consider the adjoint action of H on itself and it's elementary to show (using the adding a co(1) trick x=x_(1)S(x_(2))x_(3), which is basically g=gg^(-1)g for groups) that the center Z(H) is exactly given by invariants under the adjoint action, i.e. those elements in H, which get fixed under the adjoint action. This arises the following question (since Z(H)=HH^0(H)) : Can all Hochschild cohomologies of a Hopf algebra H be expressed in terms of the invariants functor? In particular is there a connection between things like group cohomology and Hochschild cohomology of the group algebra? Interesting stuff!
Totally different but also very nice thing for all representation theorists: If g is a Lie algebra and V is a g-rep, then the action map g otimes V -> V is a homomorphism of g-representations! Crazy stuff! In particular inserting V=g itself, one observes that the Lie bracket itself is a homomorphism of representations. Connected phenomenon: The definition of U(g) is literally: take g (which is a rep of itself), take all the tensor powers (which are again reps), direct sum them all together, giving the tensor algebra (which is a rep of g) and quotient out the two-sided ideal generated by (x otimes y - y otimes x - [x,y]). HOWEVER: Note that the thing we quotiented out is a subrepresentation of T(g) over g if you want by a consequence of Jacobi-identity! Equivalently U(g) is a g-rep, via the adjoint action! In this light the PBW theorem is just U(g) is isomorphic as g-rep via the adjoint action to Sym(g), in particular the centre of of U(g) corresponds to invariants of U(g) (under the adjoint action) and hence to invariants of Sym(g) under the adjoint action!
Ulis (my advisors) perspective on this: This at least allows to compute the size of the centre of U(g), but unfortunately doesn't give the algebra structure on it. For this we need more heavy machinery (if g is semisimple) like the Harish-Chandra isomorphism.
What do we learn from all of this: Whenever there are representations, there are even more representations one forgot to consider! Even things not looking like reps (like the center) are secretly reps!
31.03.23 (Back from argentina, center of U(sl2) and a type B q-polynomial)
I am back in Germany and first day again at the university. Flying seems like one of the craziest and at the same time worst inventions of human kind. At least from an environmental perspective. However meeting everyone in Argentina and getting new perspectives on life and math is awesome. A short but beautiful thing I learned from David Rose's lecture, was how to compute the center of U(sl2) using PBW from character theory. Roughly speaking by PBW the associated graded of U(sl2) viewed as rep of sl2 is Sym(L(2)), from which we can compute it's character (which is some polynomial in q, q^(-1) & t) and compute the multiplicity of the trivial rep in each graded component (which is some polynomial in t). Using this computation and showing that powers of the Casimir are non-zero, proves that Z(U(sl2)) is a polynomial ring in the Casmir ef + fe + h^2/2. Different but cool thing I cooked up:
The q-polynomial of the type B Weyl group is (1+q)(1+q^2)...(1+q^n)[n]'_q[n-1]'_q...[1]'_q. This is pretty wonderful since it should compute the dimensions of Schubert cells in the type B/C full flag variety! :) I will work this year with George Balla on Symplectic Grassmanians and stuff in Aachen. This will be nice!
10.03.23 (Beginning)
I decided to set up this math diary page, in case I am bored and my brain wants to put something here. Tomorrow I go for the first time a country on a different continent being Argentina for this wonderful summer school CIMPA 2023. Otherwise learning some more Hochschild stuff today I guess.
Comment added afterwards: Shoutout to the organizers, other particapants and my deer friends Timm, Daniel and Jonas who travelled with me.