User charris - MathOverflow

Character of the Basic Representation for Affine E_8 in Terms of Jacobi Theta Functions

2011-03-03T03:28:02Z

When $\mathfrak g$ is a complex, simple, simply laced Lie algebra of rank r then the (specialized) character of the basic representation for the corresponding affine Lie algebra $\hat {\mathfrak g}$ is given by $\chi_{\hat {\mathfrak g}}(q)=\frac{\Theta _{\mathfrak g}(q)}{\eta (q)^r}$, where $\Theta _{\mathfrak g} (q)$ is the theta function for the root lattice of $\mathfrak g$. It is well known that $\Theta _{E_8} (q)$ is equal to the normalized Eisenstein series of weight 4, $E_4 (q)$. We also have another basic result that $E_4(q)=\frac{1}{2}(\theta_2 (0,q)^8 +\theta_3 (0,q)^8 + \theta _4 (0,q)^8)$, where $\theta_2(x,q)$, $\theta_3(x,q)$, $\theta_4(x,q)$ are the three classical even Jacobi theta functions. Being odd, the other Jacobi theta function $\theta_1(x,q)$ vanishes at $x=0$. My question is in regards to the unspecialized form of the character; that is, I'm wondering if evaluating the character on a non-zero element of the Cartan subalgebra produces something of the form $$\chi _{\hat {E_8}}(x_1,...,x_8,q)=\frac{\frac{1}{2}(\prod_{i=1}^8\theta_2 (x_i,q) + \prod_{i=1}^8\theta_3(x_i,q) + \prod_{i=1}^8\theta _4 (x_i,q)+\alpha \prod_{i=1}^8\theta_1 (x_i,q))}{\eta (q)^8}$$ for some $\alpha$? Moreover, in the case $\alpha = 0$, does it make sense that Taylor expanding the numerator of this expression and writing it in terms of elementary symmetric polynomials in $x_1^2,...,x_8^2$ one should find that everything below degree 8 can be written as quasimodular forms times only powers of $p_1(x_1,...,x_8)=x_1^2+...+x_8^2$? This is what I got when I plugged it into Mathematica and in fact the coefficients are similar to the Eisenstein-Jacobi series $E_{4,1}(z,q)$. What would probably be very relevant is if there's anything regarding Jacobi forms in several variables that has been studied that is similar to the "development coefficients" which are well studied in Eichler and Zagier "The Theory of Jacobi Forms".

I've tried looking through a good bit of the literature regarding both of these matters, but couldn't find what I was looking for. If anyone could point me in the right direction, I would be greatly appreciative.

Getting certain modular functions from characters

2011-09-09T00:52:15Z

It is well known that characters of affine Lie algebras have certain modular properties. For instance, the linear span of all irreducible characters at a given level must be invariant under a certain action of $SL(2,\mathbb Z)$. In the case of affine $E_8$ there is only one irreducible level $1$ representation, the basic representation $V(\Lambda_0)$, and the (specialized and normalized) character can be written as $$\chi_{V(\Lambda_0)}(q)=\frac{E_4(q)}{\eta(q)^8}.$$ The RHS can be achieved as a sum of characters of another affine algebra. Affine $so(16)$ has $4$ level one representations. Besides the basic representation another one of these is $V(\Lambda_4)$, where $\Lambda_4$ denotes the fundamental weight whose finite part is the highest weight for one of the half spin representations. Using specialized and normalized characters again we have $$\chi_{V(\Lambda_0)}(q)+\chi_{V(\Lambda_4)}(q)=\frac{E_4(q)}{\eta(q)^8}.$$ I am interested in which elements of $\mathbb Z [E_4,E_6,\Delta]/(E_4^3-E_6^2-1728\Delta)$ also can show up here. It's not hard to use the above to get $\frac{E_4(q)^n}{\eta(q)^{8n}}$, so a good starting spot I'm wondering about is:

Question:is there an affine Lie algebra and a finite set of virtual representation $V_1,...,V_n$ such that $$\chi_{V_1}(q)+...+\chi_{V_n}(q)=\frac{E_6(q)}{\eta (q)^{12}}$$

The need for virtual representations is certainly necessary since the RHS will have some negative coefficients. I suspect the answer is no, because I'm guessing the whole thing is tied to even unimodular lattices and the second way above of getting $\frac{E_4(q)}{\eta(q)^8}$ comes from the connection between $E_8$ and $SO(16)$. So if not, is it possible to achieve this by some other infinite dimensional algebras whose characters have modular properties, e.g. generalized Kac-Moody algebras, vertex operator algebras, etc...

Characteristic Classes for $E_8$ Bundles

2011-11-09T03:09:28Z

Given a principal $E_8$ bundle $P\rightarrow X$ one can take the adjoint representation $\rho :E_8\rightarrow SU(\mathbb C^{248})$ and form the associated vector bundle $V=P\times_{\rho}\mathbb C^{248}$. This vector bundle has an $E_8$ characteristic classes $\lambda (V)\in H^4(X,\mathbb Z)$ obtained by pulling back the generator $\lambda \in H^4(BE_8,\mathbb Z)$ and it has a second Chern class $c_2(V)$ given by pulling back the generator $c_2\in H^4(BSU(248),\mathbb Z)$.

I am looking for a reference for the following fact: $$\lambda (V)=\frac{c_2(V)}{60}.$$

There is the map $B\rho^* : H^4(BSU(248),\mathbb Z)\rightarrow H^4(BE_8, \mathbb Z)$. Since both groups are canonically isomorphic to $\mathbb Z$, the map is determined by a single integer, which is the Dynkin index of $E_8$ and has been computed to be $60$. The above fact essentially follows from this. I am writing a paper and would prefer to just state the fact and then point the reader to whoever first presented a thorough argument filling in all the details. I have seen it mentioned as a footnote on page 68 in this paper by Friedman, Morgan, and Witten and some of the details regarding the Dynkin index are discussed in this paper by Totaro.

Answer by charris for affine Kac-Moody algebras

2011-09-15T01:30:44Z

I think you might like Affine Lie Algebras and Quantum Groups by Jurgen Fuchs. Also, Lie Algebras of Finite and Affine Type by Roger Carter is pretty good.

Example of a nontrivial fiber bundle with total space compact, spin, and $p_1=0$

2011-07-20T19:29:28Z

I'm hoping someone can help me out with finding an example of the following:

a nontrivial fiber bundle $Y \hookrightarrow Z \rightarrow X$ where $X,Y,$ and $Z$ are all compact even dimensional spin manifolds with first Pontryagin classes satisfying $p_1(Z)=0$ and $p_1(X)\neq 0$. I'd also like dim $Y\geq8$.

Thanks in advance.

Binomial coefficients and derivatives of modular forms

2011-06-28T22:32:44Z

Let $E_2$, $E_4$, and $E_6$ denote the standard Eisenstein series. The usual variables $q=e^{2\pi i\tau}$ allow us to regard the $E_n$'s as functions on either the upper half plane or the unit disk and we can define $E_n'=\frac{1}{2\pi i}\frac{d}{d\tau}E_n(\tau)=q\frac{d}{dq}E_n(q)$. I had cause to calculate a few of these and saw $$ E_4'=\frac{1}{3}(-E_6+E_4E_2) $$ $$ E_4''=\frac{5}{36}(E_8-2E_6E_2+E_4E_2^2)$$ $$E_4^{(3)}=\frac{5}{72}(-E_{10}+3E_8E_2-3E_6E_2^2+E_4E_2^3) $$ $$E_4^{(4)}=\frac{35}{864}(E_4^3-4E_{10}E_2+6E_8E_2^2-4E_6E_2^3+E_4E_2^4)-40\Delta $$ and $$E_6'=\frac{1}{2}(-E_8+E_6E_2) $$ $$E_6''=\frac{7}{24}(E_{10}-2E_8E_2+E_6E_2^2) $$ $$E_6^{(3)}=\frac{7}{36}(-E_4^3+3E_{10}E_2-3E_8E_2^2+E_6E_2^3)+168\Delta $$ It's a standard fact that the derivative of a modular form is quasimodular, so it's not surprising that we have polynomials in $E_2$. I am surprised about the appearance of the binomial coefficients though. Is there a deeper reason for their appearance? Also, I wonder if the/a pattern continues. For instance, it would be interesting if it happens that there always is some $\alpha \in \mathbb{Q}$ so that $$E_4^{(n)}-\alpha \sum_{k=0}^{n} (-1)^{k+n}\binom {n}{k}E_{4+2n-2k}E_2^{k}$$ is modular (and similarly for $E_6$). The other direction you could ask if the pattern extends is for other modular forms besides $E_4$ and $E_6$. I've taken a handful of derivatives of other Eisenstein series and saw similar results. You don't get the binomial coefficients though when you take derivatives of $\Delta$, so maybe at most something general can be said is for non-cusp forms.

Answer by charris for lowest weight representation of loop groups

2011-05-06T20:21:34Z

The formula for the invariant bilinear form is given in $(4.9.3)$ on page 64 $$\langle (x_1,\xi_1, y_1),(x_2,\xi_2,y_2) \rangle=\langle \xi_1, \xi_2 \rangle - x_1 y_2-y_1x_2$$ As I mentioned in the comments, $(9.3.7)$ becomes then $||\mu||^2-6m=2$. So your last equation would be $m=\frac{1}{3}(a^2-ab+b^2)-\frac{1}{3}$. As you said, there's no more worries about negative $m$ and as a consistency check, for $m=0$, the solutions for $[a \ \ b]$ are $[\pm 1 \ \ 0]$, $[0 \ \ \pm 1]$, $[1 \ \ 1 ]$, and $[-1 \ \ -1]$. Applying, $B$ gives you the six weights in the Weyl orbit of $-\alpha_3$ (the roots).

Answer by charris for topological actions

2011-04-27T14:40:23Z

I don't think they actually find the constant you refer to in the update. As they remark, the choice of $\omega$ which maps to $\frac{k}{8\pi^2}Tr(F\wedge F)$ under $H^4(BG,\mathbb Z)\rightarrow H^4(BG, \mathbb R)$ is only defined up to torsion. The Chern-Simons invariant depends on the choice of $\omega$. At least that's what Dan Freed seems to indicate at the end of the appendix in this paper . He also has a follow up paper which discusses more the Chern-Simons theory where $G$ isn't necessarily simply connected.

Answer by charris for Given: $\phi(s) = \sum_p \frac{\log p}{p^s}$ Why is: $\lim_{e \rightarrow 0} e \phi(1+e)$ = 1 ?

2011-04-17T02:15:31Z

(IV) on page 706 shows that $g(s)=\Phi (s)-\frac{1}{s-1}$ is holomorphic for $Re(s)\geq 1$. Then $$\lim_{\epsilon \searrow 0} \epsilon \Phi (1+\epsilon) = \lim_{\epsilon \searrow 0} \epsilon g(1+\epsilon)+1=1$$

Chern character of the index bundle for a family of Dirac operators

2011-03-31T20:08:27Z

Suppose we have a family of compact oriented even dimensional spin manifolds ${Y_x}$ parameterized by a compact even dimensional manifold $X$. The $Y_x$'s are all diffeomorphic to some $Y$, of dimension $n$, and fit together to form a fiber bundle $\pi : Z \rightarrow X$ with fiber $Y_x=\pi ^{-1}(x)$. $TZ$ has the subbundle $V:=\text{ker }\pi_*$ which is tangent to the fibers. There may be a family of coefficient bundles also and we obtain a family of twisted Dirac operators $D_x:\Gamma(S^+_x\otimes E_x)\rightarrow \Gamma (S^-_x\otimes E_x)$. The index of the family gives rise to an element $\text{ind} D \in K(X)$, which is the virtual vector bundle $[\text{ker } D_x]-[\text{coker }D_x]$ when the dimension of both spaces are constant. Finally, there is a map $\text{H} ^{*}(Z,\mathbb{R})\rightarrow \text{H} ^{*-n}(X,\mathbb{R})$ known as the Gysin homomorphism or integration over the fibers map. We'll use the latter terminology writing the map $\int_Y$ and regarding cohomology classes as living in de Rham cohomology. The Atiyah-Singer index theorem gives

$$\text{ch }(\text{ind } D)= \int _Y \hat A (V) \text{ch}(E)$$

What general results exist regarding the components of the Chern character of the index bundle, or equivalently the results of the integration over the fibers map, for twisted Dirac operators?

To illustrate, an immediate answer is that the zero cohomology (virtual rank) is the index of the Dirac operator on $Y$. A more interesting answer is that in some cases that might be all one obtains: it is a result of Borel-Hirzebruch that the signature is strictly multiplicative in all bundles where $\pi_1$ of the base acts trivially on the rational cohomology of the fibers. The signature is the index of a certain twisted Dirac operator. If we have a family of these operators such that $Z\rightarrow X$ satisfies the condition involving the fundamental group, then the strict multiplicativity gives $\text{ch}(\text{ind }D)=\int_Y \hat A (V)ch(E)=\text{sign }(Y)$. A priori one could expect higher degree cohomology classes. It seems interesting that these vanish.

If the question is too vague or broad, I would be happy knowing

Are there any instances in which there are known relations between the Chern character of the index bundle and the Chern classes of $X$?

Comment by charris

2012-01-10T21:46:52Z

Thanks! For the second part, even with $\alpha=1$ Taylor expanding gives multiples of $(x_1^2+...+x_1^2)^n$ until degree $8$ in the $x_i's$. Could this have to do with the exponents of $E_8$ being $2,8,...$?

Comment by charris

2012-01-09T18:26:12Z

Very nice, thank you! Do you know of a good introduction to this Siegel-Weil way of expressing Eisenstein series in terms of theta functions. I have only heard about this in passing before and would like to have a better understanding of it.

Comment by charris

2011-07-20T23:01:48Z

Thanks! I figured this should be easy for people who know what they're doing. Is it possible to cook up an example where $p_n(Y)\neq 0$ for some $n\geq 2$?

Comment by charris

2011-07-03T00:02:54Z

Thanks for the great answer! I'm having a little trouble working out the last step. Is it as complicated as I seem to be making it?

Comment by charris

2011-06-29T00:02:47Z

Thanks for the response Ramsey. I did have that in mind for the formulas for $E_4'$ and $E_6'$. I iterated those formulas then to find the others. What I'm really missing is the cleverness :) I've heard this $D$ called the Serre derivative also.

Comment by charris

2011-06-28T22:50:58Z

yeah, that's what I was thinking, but I don't know how to use them here to say anything general

Comment by charris

2011-05-06T00:06:22Z

Why is the equation (9.3.7) as you have it? I would have thought $||\mu||^2-2mh=||\widetilde \lambda||^2$ should be $\langle \mu, \mu \rangle - 6m = \langle -\alpha_3, -\alpha_3 \rangle=2$

Comment by charris

2011-04-27T18:34:31Z

There's a typo in your update: $\omega \in H^4(BG,\mathbb Z)$. And I don't think things should be put that way. I wouldn't say you're looking for something ($a_{\omega}$) depending on $\omega$ from the outset. You're looking for something that depends on $M$ and $B$ and how $E$ is extended over $B$.. What I was getting at in my answer above is that once they choose to add the term of the form $\langle \gamma^* \omega, B \rangle$ this sets $\omega$ as an extra parameter.

Comment by charris

2011-04-27T18:19:29Z

I can't say I understand why that's necessarily the only sort of term that can be added, but it seems the most natural. To make everything work out you need something in terms of characteristic classes there. And that choice makes $S$ independent of the choice of bounding manifold.

Comment by charris

2011-04-02T01:15:22Z

Pontryagin classes only depend on the topological structure of the manifold. Is it ever possible that there could be relations between the Chern character of the index bundle and the Pontryagin classes on $X$, or am I still missing something?

Comment by charris

2011-04-02T00:53:36Z

Thanks for the great response. Sorry my question wasn't completely clear. I was wondering if there were other situations like we're discussing where something "interesting" happens like the index bundle is actually trivial. But I was very interested in a conceptual understanding in this case, so your answer should really help me out a lot. I could be wrong, but I don't think the triviality of the index bundle (immediately) follow from the H-C-S result. Multiplicativity of the signature doesn't require strict multiplicativity (that the fiber integral ends up only in degree 0).

Comment by charris

2011-04-01T14:33:33Z

That's good to know. Thank you for your response! I will read up on it some more.

Comment by charris

2011-03-29T16:35:39Z

Great, thank you for the reference. I will look into it.