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Given a real manifold $M$ with symplectic $2$-form $\omega$, one can ask whether the cohomology class $[\omega] \in H^2(M;{\mathbb R})$ lies in the image of $H^2(M;{\mathbb Z})$. If so, one can ask for a line bundle ${\mathcal L}$ with $c_1({\mathcal L}) = [\omega]$ (or even better, a connection $\alpha$ on $\mathcal L$ whose curvature $curv(\alpha)$ is $\omega$).

In the weakest definition of "geometric quantization", one puts a compatible almost complex structure on $M$ and uses it to define the pushforward of $\mathcal L$ to a point in $K$-theory. Call this $Q(M)$.

Are there examples worked out somewhere in which $H^2(M;{\mathbb Z})$ has torsion, so that $\mathcal L$ is not uniquely determined by $\omega$? Can $Q(M)$ depend on the choice of $\mathcal L$?

I don't have any very good reason for asking this, other than I've felt it to be a hole in my understanding of geometric quantization. The spaces I care about quantizing never seem to have torsion in $H^2$.

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up vote 8 down vote accepted

If L_1 and L_2 are two line bundles on a manifold $M$ that differ by torsion, then their Chern characters $$ch(L_1) = 1 + c_1(L_1) + \frac{1}{2}c_1(L_1)^2 + \cdots$$ $$ch(L_2) = 1 + c_1(L_2) + \frac{1}{2}c_1(L_2)^2 + \cdots$$ agree, if only because the right-hand sides of these formulas are taking place in $H^*(M;\mathbb{Q})$ and there $c_1(L_1) = c_1(L_2)$. If $L_1$ and $L_2$ are prequantizations of a symplectic structure on M, and if we can give M a compatible complex structure, then we have Q(L_1) = Q(L_2) by the Riemann-Roch formula $$Q(L_1) = \int ch(L_1) td(M) = \int ch(L_2) td(M) = Q(L_2)$$

We do not get a finer equation than $Q(L_1) = Q(L_2)$. Below the line is an example of $M$, $L_1$, and $L_2$ for which $H^*(M;L_1)$ and $H^*(M;L_2)$ are not isomorphic, that I thought of first. I'll integrate these answers later.

I had written an answer with a bogus example of a Kahler manifold with two prequantum line bundles $L_1$ and $L_2$ for which $Q(L_1)$ was different from $Q(L_2)$. Here's my attempt to repair it--the best I can do is a symplectic manifold with two prequantum line bundles $L_1$ and $L_2$ for which $H^0(L_1) \neq H^0(L_2)$. But there is definitely some higher cohomology that might cancel this difference when you compute $Q$. I don't have a guess for what happens.

I am using "prequantum line bundle" to mean a line bundle whose real chern class (times $2\pi i$?) is equal to the class of the symplectic form. The premise of my example is that if we have two such line bundles $L_1$ and $L_2$ whose integral Chern classes differ by $n$-torsion, then $L_1$ and $L_2$ will become isomorphic over some $\mathbb{Z}/n$-cover $Y$ of $M$. (One can see this by thinking about the bundle-theoretic meaning of the Bockstein map $H^1(M;\mathbb{Z}/n) \to H^2(M;\mathbb{Z})$ and the long exact sequence it fits into.).

We can organize the data this way: $f:Y \to M$ is a $\mathbb{Z}/n$-bundle over $M$ with a $\mathbb{Z}/n$-equivariant prequantum line bundle $L$, and $L_1$ and $L_2$ are built from $L$ by the formula $$H^0(U;L_i) = \alpha_i\text{-eigenspace of }H^0(f^{-1}(U);L)$$ where $\alpha_1$ and $\alpha_2$ are two different characters of $\mathbb{Z}/n$.

So to find such an $M$, we should find a prequantized symplectic manifold $Y$ equipped with a free $\mathbb{Z}/n$-action, such that $\mathbb{Z}/n$ does not act so uniformly on $H^0(Y;L)$ (i.e. so that the different eigenspaces of a generator for $\mathbb{Z}/n$ have different dimensions). The only source of prequantized symplectic manifolds available to me are projective varieties, and the only source of free $\mathbb{Z}/n$-actions are those on projective hypersurfaces.

So, let $Y$ be the degree 5 hypersurface in $\mathbb{P}^3$ given by the equation $x^5 + y^5 + z^5 + w^5 = 0$. If $\eta$ is a 5th root of unity, then the $\mathbb{Z}/5$-action given by $x \mapsto x$, $y \mapsto \eta y$, $z \mapsto \eta^2 z$ and $w \mapsto \eta^3 w$ is free (to check the absence of fixed points, it's important that the eigenvalues of the different letters are distinct). If $L$ is the restriction of $O(1)$ on $P^3$ to $Y$ then $H^0(Y;L) = H^0(P^3;O(1))$, which is the vector space spanned by $x,y,z,$ and $w$, and the generator of $\mathbb{Z}/5$ acts by $x \mapsto x$, $y \mapsto \eta^{-1} y$, $z \mapsto \eta^{-2} z$, and $w \mapsto \eta^{-3}w$.

The conclusion is that if $L_1$ is the line bundle on $M = Y/(\mathbb{Z}/5)$ corresponding to the eigenvalue 1 (or $\eta^{-1}$ or $\eta^{-2}$ or $\eta^{-3}$) then $H^0(M;L_1)$ is one-dimensional, and if $L_2$ is the line bundle on $M$ corresponding to the eigenvalue $\eta^{-4} = \eta$ then $H^0(M;L_2)$ is zero-dimensional.

More generally one can take (degree n hypersurface in $\mathbb{P}^r$)/(action of $\mathbb{Z}/n$), but to make the action free one needs $n > r+1$ (and maybe $n$ prime). Taking large $n$ like this creates cohomology in $H^{r-1}(M;L_1)$ and $H^{r-1}(M;L_2)$.

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The feeling I take away from this is that maybe the most satisfying answer to this question would be that one should quantize the line bundle on the cover instead, and remember that it has a $\mathbb{Z}/n\mathbb{Z}$-action... I guess it would still be good to know whether the eigenspaces had any relation to each other or not (my intuition, which seems to be the same as yours, is that they don't necessarily). –  Ben Webster Apr 28 '10 at 23:06
Our intuition is wrong! The alternating sum of the eigenspaces does not depend on the eigenvalue, so that is some kind of relation. –  David Treumann Apr 29 '10 at 16:29
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It is the group of periods of a closed 2-form $\omega$ which plays a role on the different quantizations. Every closed 2-form $\omega$ on a manifold $M$ (more generally on a diffeological space) is the curvature of a connexion on an integration bundle, a principal bundle with group the torus of periods $T_\omega = {\bf R}/P_\omega$, where $P_\omega$ is the group of periods $$ P_\omega = \{ \int_\sigma \omega \mid \sigma \in H_2(M,{\bf Z})\} \subset {\bf R}. $$ The different integration structures (bundle + connexion) are classified by $H^1(M,T_\omega)$, the different integration bundles only are classified by ${\rm Ext}(H_1(M,{\bf Z}),P_\omega)$. The torsion plays a role at the level of $H_1(M,{\bf Z})$. For example, if this group has no torsion and $M$ is compact then the integration bundle is unique, up to an equivalence.

The torus of periods $T_\omega$ is a priori equipped with the quotient diffeology of $\bf R$. Of course $T_\omega$ is a manifold only if the group of periods is generated by one number (we say that $\omega$ is integer), that is, $P_\omega = a{\bf Z}$ one says then that $\omega$ is quantizable if $a$ is a multiple of $\hbar$, $a = k\hbar$, $k \in {\bf Z}$ (it's just a normalization). But note that the construction of the integration structure doesn't need $\omega$ to be integer.

Actually this proposition is true in general if the group of periods is a strict subgroup of $\bf R$, which is always the case for a second countable manifold, but it may be not happen if $M$ is just a general diffeological space, or some very special manifold.


  1. La trilogie du moment, Annales de l'Institut Fourier, t. 45, n°3 (1995)

  2. Diffeology, in Mathematical Surveys and Monographs, 185, AMS Providence RI, (2013).

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