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In my Global Analysis course we are studying the symbols of differential operator. We did the example of the Laplacian $\Delta = dd^* + d^*d$ but there is something I do not really understand. Let me explain.

For $d: \Omega^k(M) \to \Omega^{k+1}(M)$ it is not too hard to show that, for $\xi \in T^*M$ and $\alpha \in \Lambda^kT^*M$ we have that $$\sigma(d)(\xi)\alpha = i \xi \wedge \alpha \in \Lambda^{k+1}T^*M,$$ where $\sigma(d)$ stands for the symbol of the operator $d$. Now we know that $$\sigma(d^*)(\xi) = \sigma(d)^*(\xi):\Lambda^{k+1}T^*M \to \Lambda^{k}T^*M$$ for $\sigma(d)^*$ the hermitian dual of $\sigma(d)$. From there my teacher deduced that, for $\omega \in \Lambda^{k+1}T^*M$, $$\sigma(d)^*(\xi)\omega = -i \iota_{\xi_\sharp}(\omega)$$ where $\xi_\sharp\in TM$ is the metric dual of $\xi$, i.e. $$g(\xi_\sharp, v) = \xi(v), \quad \forall v \in TM$$ and $\iota_{\xi_\sharp}(\omega)$ is the contraction $$\iota_{\xi_\sharp}(\omega) = \omega(\xi_\sharp, \cdot, \ldots, \cdot).$$ Therefore, we can compute the symbol of $\Delta$ as \begin{align} \sigma(\Delta)(\xi)\alpha &= \sigma(d) \circ \sigma(d^*)(\xi)\alpha + \sigma(d^*) \circ \sigma(d)(\xi)\alpha\\ &= \xi \wedge \iota_{\xi_\sharp}(\alpha) + \iota_{\xi_\sharp}(\xi \wedge \alpha)\\ &= \xi \wedge \iota_{\xi_\sharp}(\alpha) + |\xi|^2 \alpha - \xi \wedge \iota_{\xi_\sharp}(\alpha)\\ &= |\xi|^2\alpha \end{align} so that $\sigma(\Delta)(\xi) = |\xi|^2 \text{Id}_{\Lambda^{k}T^*M}.$ Now the thing is that I really don't understand how to deduce the form of $\sigma(d)^*(\xi)$, could one of you explain how he got this ?

Falcon
  • 4,433

2 Answers2

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The basic result one uses is the fact that the dual of the exterior product is the contraction map. More precisely, let $V$ be an $n$-dimensional complex vector space with a Hermitian metric $g$ and let $\xi \in V^{*}$ be a linear functional. Fix $k \geq 0$ and consider the map $T_{\xi} \colon \Lambda^k \left( V^{*} \right) \rightarrow \Lambda^{k+1} \left( V^{*} \right)$ given by $T_{\xi} \left( \alpha \right) = \xi \wedge \alpha$. Then the Hermitian dual $T^{*}_{\xi} \colon \Lambda^{k+1} \left( V^{*} \right) \rightarrow \Lambda^{k} \left( V^{*} \right)$ of $T_{\xi}$ (with respect to the natural induced metrics on the exterior powers) is given by $T_{\xi}^{*} \left( \omega \right) = \iota_{\xi_{\sharp}} \left( \omega \right)$. That is, we have

$$ \left< T_{\xi} \left( \alpha \right), \omega \right>_{\Lambda^{k+1} \left( V^{*} \right)} = \left< \xi \wedge \alpha, \omega \right>_{\Lambda^{k+1} \left( V^{*} \right)} = \left< \alpha, \iota_{\xi_{\sharp}} \left( \omega \right) \right>_{\Lambda^k \left( V^{*} \right)} = \left< \alpha,T_{\xi}^{*} \left( \omega \right) \right>_{\Lambda^{k} \left( V^{*} \right)} $$

for all $\alpha \in \Lambda^k (V^{*}), \omega\in \Lambda^{k+1} (V^{*})$.

To see this, note that both sides are multilinear with respect to $\xi,\alpha,\omega$ so it is enough to verify the identity on a basis. Choose an orthonormal basis $e_1,\dots,e_n$ for $V$ and let $e^1,\dots,e^n$ be the dual basis. Let $1 \leq i_1 < \dots < i_k \leq n$ and $1 \leq j_1 < \dots < j_{k+1} \leq n$. Then $e^i_{\sharp} = e_i$,

$$ \left< e^i \wedge \left( e^{i_1} \wedge \dots \wedge e^{i_k} \right), e^{j_1} \wedge \dots \wedge e^{j_{k+1}} \right> = \begin{cases} 0, & \{i, i_1, \dots, i_k\} \neq \{ j_1, \dots, j_{k+1} \} \\ (-1)^l & \{i, i_1, \dots, i_k\} = \{ j_1, \dots, j_{k+1} \} \textrm{ and } i= j_{l+1} \end{cases}$$ while $$ \left< e^{i_1} \wedge \dots \wedge e^{i_k}, \iota_{e_i} \left( e^{j_1} \wedge \dots \wedge e^{j_{k+1}} \right) \right> = \sum_{l=0}^k (-1)^l \left< e^{i_1} \wedge \dots \wedge e^{i_k}, e^{j_1} \wedge \dots \wedge e^{j_l} \wedge \iota_{e_i} e^{j_{l+1}} \wedge \dots \wedge e^{j_{k+1}} \right>$$ which is readily seen to be the same expression (as $\iota_{e_i} e^j = \delta_i^j$).

dejavu
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A few steps:

  1. Fix a 1-form $\eta$. Using metric $g$, show that any $k$-form $\beta$ has a unique decomposition $\beta=\eta\wedge \beta'+\beta''$ (think of orthogonal decomposition, see also item 2).

  2. Show that given any other $(k-1)$-form $\alpha$, $g(\eta\wedge \alpha, \beta'')=0$, and $g(\eta\wedge\alpha, \eta\wedge\beta)=|\eta|^2g(\alpha,\beta')$.

  3. Show that $\iota_{\eta_\#}(\beta)=|\eta|^2\beta'$ (a local coordinate computation).

In the end, show that $g(\eta\wedge \alpha, \beta)=g(\alpha, \iota(\eta_{\#})\beta)$, which (up to rescaling by $\pm i$) gives the adjunction we want.

Bobby123
  • 36