# A warmup: GL_t for t not an integer

This post is meant as a warm-up to my planned follow-up to David’s post.  You don’t have to have read his post to understand this one, but there are a few technical details at the end where I’ll refer you to the end of his post.  Most of what I learned here I learned from reading this expository paper by Ostrik which I read in preparation for some talks I gave my second year of grad school.

If you like to draw pictures, how do you think about the representation theory of groups?  Well, you use an oriented strand for some basic or fundamental representation V of a group, you orient the strand the other way for the dual representation, you use disjoint union for tensor product.  Now you can try to draw pictures for maps between tensor products $V\otimes V \otimes V^* \rightarrow V^* \otimes V^*$ say.  Stacking these pictures is composition, and disjoint union is tensor product.  This should be pretty familiar to you if you’ve read the archives for “this week in mathematical physics.”

Since we’re looking at the category of representations of a group we have a bonus bit of information: this tensor category is symmetric.  There’s a canonical map $V\otimes W \rightarrow W \otimes V$ which satisfies the relations of the symmetric group.  In pictures this can be drawn using a crossing.  (Warning: this is not a crossing in 3-dimensional space, you need to either think of your pictures as being in 4-dimensions or not embedded at all.)

Ok, so what is $\mathrm{GL}_t$?  It should be a linear symmetric tensor category with a representation V that has no properties other than having dimension t.  What does it mean to have dimension t in picture language?  It means that a closed loop should have the value t.  So here’s our proposed category $\mathrm{GL}_t$:

• Objects are collections of oriented points on a line.
• Morphisms are linear combinations of oriented strands (unembedded or in dimension greater than 4 so that the crossings satisfy the relations of the symmetric group) whose boundaries match the objects that they’re mapping between.
• Composition is stacking of diagrams with the relation that a closed loop can be removed for a multiplicative factor of t.
• Tensor product is disjoint union.

Here’s a typical morphism in this category: Ok, if t is an integer have we recovered the usual category of representations of $\mathrm{GL}_n$?  Well certainly not, because the only objects are tensor products of V and its dual, we don’t have anything like Sym^2 V.  This isn’t a big problem, since V is faithful, every representation occurs as a summand in some tensor power of it.  Hence to recover all the reps we need only formally adjoin images of all projections.  This is called taking the idempotent completion, or the Karoubi envelope, and David’s post explains it nicely.

The second problem is a bit more subtle. We also want to kill all “negligible morphisms” (this is another way of saying kill the radical as in David’s post).  A negligible morphism is a morphism that when you close it off in any way (thereby getting an endomorphism of the trivial object) you always get zero.  The cool thing about the collection of negligible morphisms is that its the unique maximal planar ideal!  If you’re going to kill any morphism you have to kill off a negligible morphism (otherwise you’d kill its closing off which would kill a nonzero multiple of the empty diagram and hence kill everything), and you can kill of all the negligble morphisms if you want to.  Now why would you want to?  Well for one thing if you want your category to be semisimple (as we know the category of representations of a reductive group must be) then you’d better kill the negligible morphisms.  So killing off the negligible morphisms means we’re not just looking at any n-dimensional representation of a group, we want to make sure that we’re looking at a representation of a reductive group.

This is a little subtle.  Can anyone come up with a reason to study GL_t for t an integer but without killing off the negligibles?  Might it be useful say in studying the representation theory of the group of upper-triangular matrices?  Or the supergroup GL(m|n) with m-n=t?  I’m not sure.  Is there a better reason than semismplicity (which is not an obvious property of GL_n) to explain why you should kill off all neglibiles?

And now here’s the miracle, once you fix these two technical issues you really do recover GL_n.  It should be possible to give some category theoretical proof of it.  Clearly there’s a functor from this GL_n to the usual category of representations of GL_n.  So if you could show that this category really is the category of representations of a group then you could use the universal property of GL_n as the largest group acting on an n dimensional vector space to show that this functor is an equivalence.  This last step requires using some more heavy-duty results of Deligne that I’m not going to get into.

I’ll finish this off with a few exercises:

1. What’s the relationship between the above and Schur-Weyl duality?
2. What about the orthogonal group O_t?  What’s the only new property that the standard representation of the orthogonal group has?  How would you write that in terms of pictures.
3. Do all the above for quantum gl_t.  (Hint: HOMFLY!)
4. Interpret the above in terms of the representation theory of GL_n for n really large.  Which kinds of stable representations are you talking about?  What’s the combinatorics for describing them?  (The answer to this explains the right combinatorics for simple objects in GL_t for t generic.)

## 20 thoughts on “A warmup: GL_t for t not an integer”

1. Is there a better reason than semisimplicity (which is not an obvious property of GL_n) to explain why you should kill off all negligible?

Isn’t “all neglible morphisms are trivial” just a restatement of one of the basic theorems of multilinear algebra (the space of invariants of a big tensor product of V’s and V^*’s is spanned by all ways ways of pairing these)? That seems like relatively little input from multi-linear algebra, but maybe even that’s too much?

Of course, if you want to be cute, you’ll say “neglible morphisms are unphysical.” I guess that won’t get you far.

2. Sure it’s not hard to prove (though neither is it all that easy), but I don’t want to have to reprove that result for GL_t, for O_t, for S_t, for quantum gl_t, etc. Its an annoying gap in the motivation.

3. I strongly suspect that the existence of supergroups is to blame for the neccesity of killing the negligibles. That is to say, I think the finite dimensional super-representations of GL(m+n|m), for n fixed, will all give you quotients of the diagramatic category with loop value n, but where you’ve killed off different negligible ideals. Only when m=0 do you recover the full semisimple quotient. I poked around for a while tonight trying to find something like this online, but failed. Anyone know more about supergroups than me want to clue me in?

(Incidentally the one collection of supergroups for which the representation theory is actually semi-simple, and hence where you’ve taken the full quotient by negligibles, is osp(1|2n). Thus the kill all negligibles quotient of O_t has a nice description for all integers: for positive integers it is O_n, for negative even integers it is SP_n, and for negative odd integers it is OSP(1|2n).)

4. Are there any values of $t$ other than the nonnegative integers for which you get a nontrivial radical?

5. Negative integers t (where its GL(0|n)). That should be it.

6. Huh. Computations suggest that you are correct. I can prove this in some special cases ( $Hom(V^{\otimes 3}, V^{\otimes 3})$, $Hom(\bigwedge^k V, \bigwedge^k V)$, $Hom(S^k V, S^k V)$ by using lots of nice facts about the symmetric group. Do you have a conceptual proof?

7. I’d like a more conceptual proof, but I don’t think that I have one. In particular, I’d like to know whether the same result holds in the quantum case, does U_q(GL_t) have a radical only when t is an integer? In the t integral case it also has a further radical when q is a root of unity, do these cases generalize to families where t is close to an integer and q is close to a root of unity?

The way I’d try to prove this would be (following the standard approach to the unitarity of TL) to write down the inner product matrix for each endomorphism space in terms of the diagrammatic basis. This looks kinda bad, but every individual inner product is just going to be a power of t, and the largest powers of t will be on the diagonal. We then just want to find for which values of t one of these matrices is singular. Hopefully some sort of induction will be possible here.

Also, I thought that Knop’s paper’s condition on singularity might be helpful, but didn’t fully understand it.

8. Can you get U_q(GL_t) by a Drinfeld double construction?

9. All right, I have a conjectured combinatorial proof. (No ideas for a conceptual one.)

For a permutation $w$, define $c(w)$ to be the number of cycles in $w$. Define an $n! \times n!$ matrix, with rows and columns indexed by $S_n$, whose $(u,v)$ entry is $t^{c(u^{-1} v)}$. Our goal is to compute the determinant of this matrix, and show that all the roots are integers between $-n$ and $n$.

The approach I came up with is to consider the element $z:= \sum_w t^{c(w)} w$ in the group ring of $S_n$. The matrix in question is just the action of $z$ on the regular representation. So we can instead consider the action of $z$ on all the irreducible representations of $S_n$, and multiply these together with appropriate exponents to get the determinant we care about. Since $z$ is in the center of the group algebra, it acts by a scalar on each irrep; explicitly, that scalar is $(1/\chi(1)) \sum_{w \in S_n} \chi(w) t^{c(w)}$, where $\chi$ is the character. So we just need to compute these polynomials.

For example, if $\chi$ is the trivial representation of $S_3$, we get $t^3+3t^2+2t=t(t+1)(t+2)$.

Here’s some data:

(1) t

(2) (t+1)*t
(1,1) t*(t-1)

(3) t*(t+1)*(t+2)
(2,1) (t+1)*t*(t-1)
(1,1,1) (t-2)*(t-1)*t

(4) t*(t+1)*(t+2)*(t+3)
(3,1) (t-1)*t*(t+1)*(t+2)
(2,2) (t-1)*t^2*(t+1)
(2,1,1) (t-2)*(t-1)*t*(t+1)
(1,1,1,1) t*(t-1)*(t-2)*(t-3)

I think the pattern is obvious — prod (t+i-j), where the product is over boxes (i,j) in the partition. Does anyone remember enough $S_n$ representation theory to see why?

10. So at some point you’ve reduced yourself to only caring about things of the form $\mathrm{End}(V^{\otimes k})$? That’s why you can identify the pictures with permutations, right? Is it clear why you can reduce to that case?

11. Oh, if you think of the pictures as living in circles instead of rectangles, then there will always be k inward arrows and k outward arrows, so I can think of them as just giving a permutation where the ordering is along the boundary circle. So now I believe you.

12. ” I’d like to know whether the same result holds in the quantum case, does U_q(GL_t) have a radical only when t is an integer? In the t integral case it also has a further radical when q is a root of unity, do these cases generalize to families where t is close to an integer and q is close to a root of unity?”

It seems like you must be able to perturb these roots. At the end of the day, for each $k$, the computation outputs a certain polynomial in $t$ and $q$ such that there is a nontrivial radical in $End(V^{\otimes k})$ if and only if that polynomial vanishes. Whatever the formula for that polynomial may be, the implicit function theorem tells you that if you move $q$ a little, you should be able to move $t$ a little to compensate.

13. Ben Wieland says:

Is the quantity that David isolated the dimension of the corresponding irreducible representation of GL_t? (up to a factor independent of t but dependent on the partition) It looks plausible from the determinant expression of Schur functions. Since the generic case is semisimple, the only way it can degenerate is if the identity of some element has trace zero; but the trace of the identity is the dimension.

This required input of representation theory, rather than just the cobordism category side, but I imagine the same works for quantum groups, with q-integers.

14. Darn it Ben, you’re faster than I am! This is what’s going on. In particular, there is a well known formula for the dimension of the corresponding GL_t rep, and that formula does have the property that all the roots are at integers.

15. As Ben Wieland suggests, we get nontrivial radical only when we have an irrep of dimension zero. Let me add a few details:

As sketched above, we can reduce to the problem of computing the radical of $End(V^{\otimes k})$. As a ring, $End(V^{\otimes k})$ is obviously isomorphic to $\mathbb{C}[S_k]$. By basic representation theory, $\mathbb{C}[S_k] \cong \bigoplus Mat(d_{\lambda} \times d_{\lambda})$, where $Mat(d \times d)$ is the $d \times d$ matrices and $d_{\lambda}$ are the dimensions of the irreps of $S_k$.

Now, trace function $T$ on $End(V^{\otimes k})$ has the property that $T(fg) = T(gf)$. Therefore, we must have $T=\sum a_{\lambda}(t) \mathrm{Tr}_{\lambda}$, where $\mathrm{Tr}_{\lambda}$ is the ordinary trace on $Mat(d_{\lambda} \times d_{\lambda})$. Our goal is to figure out when the $a_{\lambda}(t)$ are zero.

Now, there are two crucial observations. First, $a_{\lambda}(t)$ will be a polynomial in $t$. And, second, when $t$ is a positive integer, then $a_{\lambda}$ is the dimension of the corresponding representation of $GL_t$.

The $a_{\lambda}(t)$ are essentially what I was computing, except that for some reason I was computing $d_{\lambda} a_{\lambda}(t)$. And my conjectured formula above is a classical result; see, for example, Fulton and Harris Exercise 6.4.

$latex a_i(t) 16. The beginning of this comment which discussed the Math Jobs Wiki is now here. -Noah Okay, some obligatory math. There is a naive but somewhat valid reason to quotient by the negligible morphisms. Namely, they cannot affect the values of scalar quantities, which is often what you care about the most. Indeed, given any multiplicative invariant of anything such as graphs or links or manifolds, you can always construct a tensor category from it (which may however have infinite-dimensional hom spaces). In this reverse construction, the tensor category never has negligible morhpisms. Indeed, the situation is both related and analogous to the process of quotienting a vector space with a symmetric bilinear form, by the kernel of its bilinear form. Obviously sometimes you want to do that. The motivation is a bit naive because, for instance, you can sometimes work around a negligible morphism by considering invariants of (1,1)-tangles instead of invariants of knots. Even so, the motivation has some truth to it. Exercise: Consider the chromatic polynomial of all graphs (not just the planar ones). What are the dimensions of the hom spaces of the associated category? They are related to GL_t. 17. I’m not sure about people who read things on RSS, but my guess is that you’re more likely to get the people reading that aside if you put it in the requests thread rather than here. I know I often don’t look at comments 16 deep in a thread I haven’t been following, but always look at new things in the requests thread. Plus requesting a post is basically what you’re doing, so it’s even on-topic there. If you decide to repost it there, then I’ll delete the first half of your comment here. 18. This is a little late, but I think one reason not annihilating the negligibles may be useful is that the families of categories $\tilde{Rep}(S_t), \tilde{Rep}(GL_t)$ (which contain negligibles for$t \in \mathbb{Z}_{\geq 0}$) form a nice algebro-geometric family. (I think Pavel Etingof has used the phrase “flat family.”*) The categories without negligibles can be obtained from an internal category (to the category of $\mathbb{A}^1_{\mathbb{C}}$-schemes) by taking fibers at$t \in \mathbb{C}\$. This implies by Chevalley’s theorem that Deligne’s category $Rep(GL_t)$ must generically look like the classical categories (in particular, are generically semisimple). This has of course been proven by Deligne (who obtains a more precise result), but in the case of categories *built* out of Deligne’s categories with additional structure, for which the techniques in Deligne’s paper (at least to my knowledge, generally break down), the ideology of internal categories may still be applicable, and in particular rigorously show (as one would expect) that such families of categories behave at transcendental parameters similarly to classical representation theory.
I have written a paper on this, at
http://arxiv.org/abs/1006.1381.

Admittedly though, the non-quotiented categories are useful only as an auxilliary tool to make this approach meaningful, since the approach only yields results for $t$ transcendental, where there are no negligibles.

*I don’t know whether if in the internal category method, the associated schemes are actually flat over $\mathbb{A}^1_{\mathbb{C}}$ though.