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Let $G$ be a compact Lie group with complexification $G^c$, and consider a principal $G^c$-bundle $P^c \to M$ together with a reduction $P \subseteq P^c$ to $G$. Assume that $M$ is a Riemann surface. Then every connection on $P$ gives a unique holomorphic structure on $P^c$, and accordingly we get a natural action of the gauge group $Gau(P^c) = \Gamma(P^c \times_{G^c} G^c)$ on the space $C(P)$ of connections of $P$. One may think of $Gau(P^c)$ as the complexification of the gauge group $Gau(P)$ of $P$.

Question: Is there a relation between the stabilizer $Gau(P^c)_A$ and the stabilizer $Gau(P)_A$ of a Yang-Mills connection $A$ on $P$? At least on the level of their Lie algebras?


Background: According to Atiyah-Bott, the Yang-Mills functional can be viewed as the momentum map squared of the $Gau(P)$-action on $C(P)$ with respect to a natural symplectic structure. On the other hand, in Hessians of the Calabi Functional and the Norm Function it was shown that the Hessian of the momentum map squared yields a pair of commuting operators on the complexified Lie algebra. These operators then give a relation between the stabilizers of the complexified action and the original action. Applying this to the above Yang-Mills setting yields the following conjecture: For a Yang-Mills connection $A$, the Lie algebra $gau(P^c)_A$ of $Gau(P^c)_A$ decomposes as $$ gau(P^c)_A = \bigoplus_{\lambda > 0} E_\lambda \, \, \oplus (gau(P)_A)^{c}, $$ where $E_\lambda$ are $\lambda$-eigenspaces of $2 i [\star F_A, \cdot]$ and $(gau(P)_A)^{c}$ is the complexification of the Lie algebra of the stabilizer $Gau(P)_A$.

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  • $\begingroup$ It is not true that a connection on $P$ defines a holomorphic connection on $P_{\mathbb C}$. Take for instance $P= \mathrm{U}(n)$ and $P_{\mathbb C} =\mathrm{GL}(n,\mathbb C)$. Then you are looking at a complex Hermitian vector bundle. A Hermitian connection defines a $\bar \partial$-operator, hence a holomorphic structure, but not a holomorphic connection (which would be flat on a Riemann surface). $\endgroup$
    – Nicolast
    Commented Feb 16, 2021 at 10:30
  • $\begingroup$ Thanks for pointing this out. The bijection is of course between connections on $P$ and holomorphic structures on $P^c$. $\endgroup$ Commented Feb 18, 2021 at 14:21
  • $\begingroup$ I am not sure I understand your notations but I think $\mathrm{Gauge}(P^{\mathbb C})_A = (\mathrm{Gauge}(P)_A)^{\mathbb C}$. The case of $G=\mathrm{U}(n)$ can be worked out by decomposing your vector bundle into a direct sum of stable bundles, but probably also follows from some abstract nonsense in representation theory. $\endgroup$
    – Nicolast
    Commented Feb 19, 2021 at 15:23
  • $\begingroup$ Why do you believe that they are equal? I outlined a proof of my initial conjecture below, but maybe I've overlooked something. I would appreciate if you could have a look, thanks! $\endgroup$ Commented Feb 24, 2021 at 21:45

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For a Yang-Mills connection $A$, one indeed has a decomposition $$H_A\bigl(Ad P \otimes \mathbb{C}\bigr) = \bigl(gau(P)_A\bigr)_{\mathbb{C}} \oplus \bigoplus_{\lambda > 0} H_A\bigl(Ad_\lambda P\bigr),$$ where $H_A$ denotes the space of holomorphic sections (with respect to the holomorphic structure induced by the connection $A$) and $Ad_\lambda P$ are the eigenspaces of the endomorphism $2 i \, [\star F_A, \cdot]$ on $Ad P$.

In fact, this follows from observations made by Atiyah and Bott in their paper "The Yang-Mills equations over Riemann surfaces". On page 556 they show that the endomorphism $2 i \, [\star F_A, \cdot]$ on $Ad P$ has constant eigenvalues and thus $Ad P \otimes \mathbb{C}$ decomposes into eigenbundles $Ad_\lambda P$ for $\lambda \in \mathbb{R}$. This decomposition then induces decompositions on the level of differential forms: \begin{equation} \Omega^k(M, Ad P \otimes \mathbb C) = \bigoplus_{\lambda} \Omega^k(M, A_\lambda P). \end{equation} As a consequence of the Yang-Mills equation, the operators $\bar{\partial}_A$ and $2 i \, [\star F_A, \cdot]$ commute. Hence, \begin{equation} H_A\bigl(Ad P \otimes \mathbb C\bigr) = \bigoplus_{\lambda} H_A\bigl(Ad_\lambda P\bigr) \, . \end{equation} By considering appropriate Laplacian operators, one can show that $H_A\bigl(Ad_\lambda P\bigr)$ is isomorphic to $\bigl(gau(P)_A\bigr)_{\mathbb C}$ for $\lambda = 0$ and is trivial for $\lambda < 0$, see Lemma 5.9 (iii) and p. 559 in Atiyah and Bott.

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  • $\begingroup$ Oh ok, I understand your notations and I think it is correct. At the end you mean "$H_A(Ad_\lambda P) = (gau(P)_A)_{\mathbb C}$ for $\lambda = 0$" right ? I was thinking of a polystable bundle, for which $\star F_A = \alpha \mathrm{Id}$ commutes with everyone, so that there is no non-zero eigenspace. $\endgroup$
    – Nicolast
    Commented Feb 25, 2021 at 10:58
  • $\begingroup$ Thanks, indeed there was a $\lambda = 0$ missing. $\endgroup$ Commented Feb 25, 2021 at 11:46

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