What is the adjoint of the connection operator on a Clifford bundle?

Question

From Elliptic Operators, topology and asymptotic methods, John Roe, page 43-45.

Let $M$ be a Riemannian manifold. Let $S$ be a Clifford bundle over $M$, such that each $S_{m}$ over $m\in M$ is a left module over $Cl(\mathbb{T}_{m}M)\otimes \mathbb{C}$. Then we can define a connection on $S$ such that $$ (v\cdot s_1, s_2)+(s_1, v\cdot s_2)=0, \nabla_{X}(Ys)=(\nabla_{X}Y)s+Y(\nabla_{X}s) $$ where $s_{1},s_{2},s$ are sections of $S$, $v\in T_{m}$, $X,Y$ are vector fields.

I am confused with this definition because here $(, )$ is an inner product on $S_{m}$, and it is not defined explicitly. However, this definition still makes sense. Now John Roe derived the Dirac operator in a local orthonormal basis $e_{i}$ of $TM$ as $$ Ds=\sum_{i}e_{i}\nabla_{i}s $$ He claimed that we have the following well known Weitzenbock formula: $$ D^{2}s=-\sum_{i}\nabla^{2}_{i}s+\sum_{j<i}e_{j}e_{i}(\nabla_{j}\nabla_{i}-\nabla_{i}\nabla_{j})s\rightarrow D^{2}s=\nabla^{*}\nabla s+Ks $$ where $\nabla^{*}:C^{\infty}(T^{*}M\otimes S)\rightarrow C^{\infty}(S)$ is defined by $$ \nabla^{*}(dx^{j}\otimes s_{j})=-\sum_{k}g_{jk}(\nabla_{j}s_{k}-\Gamma^{i}_{jk}s_{i}) $$ I am very confused why this is the adjoint to the Clifford connection $\nabla$ defined earlier. It is not clear to me how the index $i$ appeared at here (implicitly using Einstein summation?), for example. Another issue is the general section of $C^{\infty}(T^{*}M\otimes S)$ should be $$ dx^{i}\otimes s_{j}, i\in \{1,\cdots n\}, j\in \{1,\cdots m\},\dim(T^{*}M)=n,\dim(S)=m $$ and I do not see any reason why he only considers $dx^{j}\otimes s_{j}$ at here. But cast this aside for now. His strategy for the proof goes as follows:

1. We want to show $$ ( s,\nabla^{*}\phi )=( \nabla s,\phi), \phi=dx^{j}\otimes s_{j}\in C^{\infty}(T^{*}M\otimes S), s\in C^{\infty}(S) $$
2. We compute by plug in the coordinates explicitly. We have $$ (s,\nabla^{*}\phi)-(\nabla s,\phi)=\sum_{k}(-g^{ik}(s,\nabla_{j}s_{k})+g^{jk}\Gamma^{i}_{jk}(s,s_{i})-g^{jk}(\nabla_{j}s, s_{k}))=\sum_{k}(-g^{jk}\partial_{j}(s,s_{k}+g^{jk}\Gamma^{i}_{jk}(s,s_{i}))=d^{*}\omega $$
3. Because the difference is a `divergence', the two sides are in fact equal.

I have some questions regarding his approach. It is not clear to me that once the difference between the two quantities is the result of first order differential operator order on a one form, then it must be zero in some sense. I suppose John Roe meant implicitly that they are equal in De Rham cohomology. It is also not clear to me that we have $$ (s, -\sum_{k}g_{jk}\nabla_{j}s_{k})=-\sum_{k}g^{jk}(s,\nabla_{j}s_{k}) $$ because the inner product on $S_{m}$ has nothing to do with the metric. And even if $(s_{i},s_{j})=g_{ij}$, I still do not see why $$ (s,g_{jk}\nabla_{j}s_{k})=g^{jk}(s,\nabla_{j}s_{k}) $$ as the inner product should be linear(or conjugate linear) with respect to either variables. It is also quite obscure to me that $$ \sum_{k}g^{jk}(\nabla_{j}s, s_{k})=(\nabla s, \phi) $$ For we have $$ (\nabla s, \phi)=(\nabla s, dx^{j}\otimes s_{j}) $$ and I do not see how he associated $s_{j}$ to $s_{k}$ by $g_{jk}$ at all.

Answer 1

Roe's book is one of my favorite books, and the place where I first learned some index theory. I know that the lack of a few details can be frustrating sometimes, though. Let me see if I can address some of your questions.

$(,)$ is the Hermitian metric on $S$ that Roe refers to in Definition 3.4. You're right that he's not too explicit about this. (I suppose since $S$ has a Hermitian metric, it must be a complex bundle.)

Roe calls $dx^j \otimes s_j$ (sum on $j$ implied) a "general section" of $T^\ast M \otimes S$. Note that he has not specified a local frame for $S$, only for $T^\ast M$. $s_j$ is not a basis element, it's a general section of $S$ that is the $S$-coefficient on the basis element $dx^j$ of $T^\ast M$.

In the proof of the formula for $\nabla^\ast$, $(,)$ represents the local (i.e., pointwise) inner product on both $S$ and $T^\ast M \otimes S$, and $\langle , \rangle$ represents the global $L^2$ inner product on both bundles. (Which bundle is being referred to must be taken from context.) Recall that the global inner product is given by the integral of the pointwise inner product times the volume form (if, e.g., $s$ and $t$ are sections of $S$): $$ \langle s, t \rangle = \int_M (s,t)_p ~ \text{vol}_p .$$ Roe mentions briefly the divergence theorem at the top of page 20, and that's what he's using here. The divergence theorem states that for a one-form $\alpha$, $$\int d^\ast \alpha ~ \text{vol} = 0.$$ You should think of this as being dual to Stokes theorem. Here's a quick proof: \begin{align*} \int d^\ast \alpha ~ \text{vol} &= \int (d^\ast \alpha, 1) ~ \text{vol} \\ &= \langle d^\ast \alpha , 1 \rangle \\ &= \langle \alpha, d(1) \rangle \\ &= 0 \end{align*} since of course the constant function $1$ is closed.

As for your last issue, you are absolutely right. There seems to be a typo in the book. The formula for $\nabla^\ast $ should involve the inverse of $g$, not $g$ itself (i.e., upper indices instead of lower indices). (Note that $g$ is the Riemannian metric on $TM$. We never need to write down the metric $(,)$ on $S$ explicitly.) It should be $$\nabla^\ast \left( \sum_j dx^j \otimes s_j \right) = -\left( \sum_{j,k} g^{jk} \nabla_j s_k - \sum_{i,j,k} g^{jk} \Gamma^i_{jk} s_i \right).$$ I have explicitly written all the sums (as you guess, there is a sum on $i$). As a reality check, in these sorts of contexts, repeated indices are almost always summed over, and they almost always appear with one "up" and one "down".

Note that $\nabla s = \sum_k dx^k \otimes \nabla_k s$, so $$(\nabla s, \varphi) = \left(\sum_{k} dx^k \otimes \nabla_k s, \sum_j dx^j \otimes s_j \right) = \sum_{j,k} g^{jk} (\nabla_k s, s_j).$$ It's a bit odd that he only explicitly writes the sum on $k$, when really there are implied sums on all the indices that appear twice.

I hope that helps.

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Edit: Here I will address a question from the OP posed in the comments below.

By definition, $dx^k(\partial_i) = \delta^k_i$. This is always the case, no matter the coordinate system. That's just what $dx^k$ means as a local section of $T^\ast M$. You were also correct in writing $dx^k(\partial_i) = g^{kj}g_{ji}$, but this is precisely $\delta^k_i$, since $g^{kj}g_{ji}$ is the $(k,i)$th entry of the identity matrix $g^{-1} g$.

I think it's safe to say that almost anytime one wants to identify $TM$ and $T^\ast M$ in the presence of a Riemannian metric, one will use the musical isomorphism (without a metric, there is no canonical identification). The musical isomorphism $\sharp: T^\ast M \to TM$ is defined by $\omega(X) = (\omega^\sharp, X)$. But there's no need to use this to evaluate $dx^k(\partial_i)$, because it's always just $\delta^k_i$.

These ideas having to do with vector spaces (and bundles) and their duals are a common source of confusion. A quick search returns this SE question, but there are many other sources of course.