4
$\begingroup$

This question is mainly about Section 5.2 of the book "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne.

Let $M$ be a compact Riemannian manifold without boundary and $P\rightarrow M$ be a principal bundle with compact structure group $G$. Let $E$ be a finite dimensional complex vector space and $\rho: G\rightarrow End(E)$ be a representation, then we get an associated vector bundle $\mathcal{E}:= P\times_G E$ over $M$.

If we have a connection on the principal bundle $P$, then we have the associated connection on the associated bundle $\mathcal{E}$, then we have the Laplacian operator $\Delta^{\mathcal{E}}$, which acts on $\Gamma(M,\mathcal{E})\cong (C^{\infty}(P)\otimes E)^G$.

On the other hand, the connection on $P$, together with the Riemannian metric on $M$ and the invariant metric on $G$, gives a Riemannian metric on the bundle (considered as a manifold) $P$. therefore we have the scalar Laplacian $\Delta^P$ acting on $C^{\infty}(P)$. Furthermore $\Delta^P\otimes 1$ acts on $C^{\infty}(P)\otimes E$.

By some computation we get the Proposition 5.6 of that book: The Laplacian $\Delta^{\mathcal{E}}$ coincide with $\Delta^P\otimes 1+1\otimes \text{Cas}$ on $(C^{\infty}(P)\otimes E)^G$, where Cas is the Casimir operator on $E$.

Then we have the heat kernel $e^{-t\Delta^{\mathcal{E}}}\in \Gamma (M\times M, \mathcal{E}\boxtimes \mathcal{E}^*)$ and we can pull it back to $P$ and get a function $e^{-t\Delta^{\mathcal{E}}}\in C^{\infty}(P\times P)\otimes End(E)$.

On the other hand the scalar Laplacian $\Delta^P$ also has a heat kernel $ e^{-t\Delta^P}\in C^{\infty}(P\times P)$. The book claims that by the above Proposition 5.6 we have $$ e^{-t\Delta^{\mathcal{E}}}= e^{-t\text{Cas}}e^{-t\Delta^P}. $$

My first question is: in what sense the above equality holds? Since $\Delta^{\mathcal{E}}$ is defined only on $(C^{\infty}(P)\otimes E)^G$ and $\Delta^P$ is defined on the whole $C^{\infty}(P)\otimes E$, it's not obvious that the two heat kernels are equal as elements in $C^{\infty}(P\times P)\otimes End(E)$. My guess is that they are equal as operator semigroups acting on $(C^{\infty}(P)\otimes E)^G$.

Then comes Proposition 5.7: If $p_1$ and $p_2$ are two points on $P$, then $$ < p_1|e^{-t\Delta^{\mathcal{E}}}|p_2 >= e^{-t\text{Cas}} \int_G < p_1|e^{-t\Delta^P}|p_2 g >\rho(g)^{-1}dg. $$

The proof of Proposition 5.7 in the book is to pick a test function $\phi\in (C^{\infty}(P)\otimes E)^G$ and we know $$ e^{-t\Delta^{\mathcal{E}}}\phi=e^{-t\text{Cas}}e^{-t\Delta^P}\phi $$ Then we calculate the right hand side and get the result.

My second question is: Is it sufficient to prove the proposition using only $\phi\in (C^{\infty}(P)\otimes E)^G$? It seems that the $G$ invariant function is not enough. By a similar argument we can "prove" that $$ < p_1|e^{-t\Delta^{\mathcal{E}}}|p_2 >= e^{-t\text{Cas}}< p_1|e^{-t\Delta^P}|p_2 > $$ but this is not true in general. Then what have I ignored in this proof of Proposition 5.7?

$\endgroup$
1
$\begingroup$

Your understanding is correct. All the identify is only for the $G$-invariant section.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

About first question:

$e^{-t\Delta^\mathcal{E}} = e^{-t\mathrm{Cas}}e^{-t\Delta^P}$ is a equation between elements of $\mathrm{End}(C^\infty(P,E)^G)$. More precisely, $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1} = e^{-t\mathrm{Cas}}e^{-t\Delta^P}$, where $\alpha: \Gamma(M,\mathcal{E}) \xrightarrow{\simeq} C^\infty(P,E)^G$.

About second question:

$\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle = e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg\rangle\rho(g)^{-1}|dg|$ holds for all $p,q \in P$.

First, $\Gamma(M\times M,\mathcal{E}\boxtimes\mathcal{E}^*)) \simeq C^\infty(P\times P,\mathrm{End}(E))^{G\times G}$, where $G\times G$ acts on $\mathrm{End}(E)$ by $G\times G\times\mathrm{End}(E) \ni (g,h,A) \mapsto \rho(g)A\rho(h)^{-1} \in \mathrm{End}(E)$. $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle \in C^\infty(P\times P,\mathrm{End}(E))^{G\times G}$ is defined as the correspondent of the heat kernel $\langle x|e^{-t\Delta^\mathcal{E}}|y\rangle \in \Gamma(M\times M,\mathcal{E}\boxtimes\mathcal{E}^*))$. $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle$ is not the "kernel" of the operator $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1}$, but $|G|$ times of the kernel of the operator $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1}P_G$. Here, $P_G: C^\infty(P,E) \to C^\infty(P,E)^G$ is defined by $(P_G s)(p) = \frac{1}{|G|}\int_G\rho(g)s(pg)|dg|$.

Let's see $|G|$ times of the kernels of the equation $e^{-t\Delta^\mathcal{E}}P_G = e^{-t\mathrm{Cas}}e^{-t\Delta^P}P_G$. $|G|$ times of the kernel of the left hand side is $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle$, and that of the right hand side is $e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg^{-1}\rangle\rho(g)|dg| = e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg\rangle\rho(g)^{-1}|dg|$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.