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$\newcommand{\sig}{\sigma}$ $\newcommand{\tr}{\operatorname{tr}_{\eta}}$ $\newcommand{\al}{\alpha}$ $\newcommand{\be}{\beta}$ $\newcommand{\til}{\tilde}$

Let $E$ be a smooth vector bundle over a manifold $M$. Suppose $E$ is equipped with a metric $\eta$ and with a compatible metric connection $\nabla$. Note that $\nabla$ induces a connection on $E \otimes E$.

Let $\sig \in \Omega^k(M,E \otimes E)$ be an $E \otimes E$-valued differential form of degree $k$.

We denote by $d_{\nabla}$ the corresponding covariant exterior derviative on $E \otimes E$-valued forms:

$$ d_{\nabla}:\Omega^k(M,E \otimes E) \to \Omega^{k+1}(M,E \otimes E) $$

Now we can apply contraction and exterior differentiation in two orders:

(1) Let $\sigma \in \Omega^k(M,E \otimes E)$; $\, \,d_{\nabla} \sig \in \Omega^{k+1}(M,E \otimes E)$, so $\tr(d_{\nabla} \sig) \in \Omega^{k+1}(M)$ (a real valued form).

(2) $\tr(\sig) \in \Omega^{k}(M) \Rightarrow d\left(\tr(\sig)\right) \in \Omega^{k+1}(M)$ ($d$ is the standard exterior derivative of course)

Question: Is it true that $\tr(d_{\nabla} \sig) =d\left(\tr(\sig)\right)$ ?

I tried to prove this via induction on the degree of the form, but I am unable to finish the proof (see details below).

I suspect there is a more general commuting property hiding in the shadows, i.e commutation between exterior derivative and contraction for $E^* \otimes E$-valued forms.

I would be happy to find a reference (or a self-contained proof, of course).

Here is an attempt to prove this by induction on the degree:

$k=0$: $\sig \in \Omega^0(M,E \otimes E)=\Gamma(E \otimes E)$. Let $X \in \Gamma(TM)$.

$$ d_{\nabla} \sig \,(X)=\nabla_x^{E \otimes E} \sig \in \Gamma(E \otimes E)$$

Since the assertion is local, we can assume $\sig=\al \otimes \be$, where $\al ,\be \in \Gamma(E)$. So, on the one hand

$$ (*) \tr(d_{\nabla} \sig) (X) = \tr \left(d_{\nabla} \sig \left(X\right )\right) = \tr(\nabla_x^{E \otimes E} \left (\al \otimes \be\right))=\tr(\nabla_x^E \al \otimes \be) + \tr( \al \otimes \nabla_x^E \be)$$

$$ =\widetilde{\nabla_x^E} \al \,(\be) + \til \al (\nabla_x^E \be) $$

where given a section $\gamma \in \Gamma(E)$, we denote by $ \til \gamma \in \Gamma(E^*)$ its corresponding section obtained using the metric $\eta$.

On the other hand,

$$ \tr(\sig) =\tr(\al \otimes \be)= \til \al (\be)= \langle \al , \be \rangle_{\eta} \in \Omega^0(M)=C^{\infty}(M),$$ hence

$$ (**) \, d\left(\tr(\sig)\right) (X)= X \cdot \langle \al , \be \rangle_{\eta} = \langle \nabla_X^E \al , \be \rangle_{\eta} + \langle \al , \nabla_X^E \be \rangle_{\eta} =\widetilde{\nabla_x^E} \al \,(\be) + \til \al (\nabla_x^E \be)$$

Now $(*), (**)$ implies the desired equality.

Now, let us assume the assertion holds for all $j$-forms when $j < k$. (I am not sure how to proceed from here)

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The general version you are looking for is the following: Suppose that$\Phi:V\to W$ is a vector bundle map between two vector bundles $V$ and $W$ over $M$. Suppose that we have connections $\nabla^V$ and $\nabla^W$, which are compatible with $\Phi$ in the sense that $\nabla^V_\xi \Phi(s)=\Phi(\nabla^W s)$ for all $s\in\Gamma(V)$ and for any vector field $\xi\in{\frak{X}}(M)$. Then $\Phi$ induces maps $\Phi_*:\Omega^k(M,V)\to\Omega^k(M,W)$ by acting on the values of forms. The general statement is that $\Phi_*$ is compatible with the covariant exterior derivatives, i.e. $d^{\nabla^W}(\Phi_*(\alpha))=\Phi_*(d^{\nabla^V}\alpha)$ for any $\alpha\in\Omega^k(M,V)$. The case you are looking at is $V=E\otimes E$ and $\nabla^V$ the induced connection from the metric connection $\nabla$ on $E$ and $W=M\times\mathbb R$ with the trivial connection (so the covariant exterior derivative becomes the exterior derivative) and $\Phi=tr_{\eta}$. The compatibility of $tr_{\eta}$ with the two connections essentially is the definition of a metric connection and it is verified your question.

Now for the proof of the general version, you just need the "global formula" for the covariant exterior derivative: Take vector fields $\xi_0,\dots,\xi_k\in{\frak X}(M)$ and a form $\beta\in\Omega^k(M,W)$. Then you can write $(d^{\nabla^W}\beta)(\xi_0,\dots,\xi_k)$ as $$ \sum_i(-1)^i\nabla^W_{\xi_i}\beta(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)+\sum_{i<j}(-1)^{i+j}\beta([\xi_i,\xi_j],\xi_0,\dots,\widehat{\xi_i},\dots,\widehat{\xi_j},\dots,\xi_k) $$ with the hats denoting omission. If you insert $\beta=\Phi_*\alpha$, then in the second sum you simply get $\beta(\dots)=\Phi(\alpha(\dots))$. Doing a similar replacement in the first sum, you then use compatibility of $\Phi$ with the connections as $$ \nabla^W_{\xi_i}\Phi(\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k))= \Phi(\nabla^V_{\xi_i}\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)). $$ Hence you conclude that $(d^{\nabla^W}\Phi_*\alpha)(\xi_0,\dots,\xi_k)=\Phi(d^{\nabla^V}\alpha(\xi_0,\dots,\xi_k))$ which is exactly the claim $d^{\nabla^W}\Phi_*\alpha=\Phi_*d^{\nabla^V}\alpha$.

Edit: As mentioned in the comment, the compatibility condition should read as $\nabla^W_\xi \Phi(s)=\Phi(\nabla^V_\xi s)$ for all $\xi$ and $s$.

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  • $\begingroup$ Thanks! nice answer. I guess you have a minor typo in the statement of the compatibility condition; It is supposed to be $ \nabla_{\xi}^W (\Phi(s))=\Phi(\nabla_{\xi}^V s)$. The use of the global formula is of course very reasonable. I am still wondering however if this property can be deduced directly from the Leibnitz rule via induction (that is, from the defining property of the exterior derivative, without using the explicit formula). $\endgroup$ – Asaf Shachar Oct 24 '16 at 7:41
  • $\begingroup$ Thanks for pointing out the typo, I have edited the answer. The alternative to using the global formula is similar to the answer of @Sebastian : Write forms as $\alpha\otimes s$ for $\alpha\in\Omega^k(M)$ and $s\in\Gamma(V)$ and use that $d^{\nabla^V}(\alpha\otimes s)=d\alpha\otimes s+\alpha\wedge\nabla^V s$. But in both versions, it is not really necessary to involve an induction by the degree of forms. $\endgroup$ – Andreas Cap Oct 24 '16 at 7:56
  • $\begingroup$ I am going to publish a paper which uses vector valued-forms quite heavily. For completeness, I am reproducing your argument in the appendix (I need this commutation property as a small lemma inside a much vaster argument...). Of course, I plan to give you credit for this (at least in the acknowledgment, but I am also thinking to actually cite your answer). If you have any objections or preferences, please let me know. Thanks again for your help. $\endgroup$ – Asaf Shachar Jan 7 '17 at 12:40
  • $\begingroup$ Thanks for the plan to acknoledge the answer. I am fine with this and don't have preferences. $\endgroup$ – Andreas Cap Jan 7 '17 at 15:23
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As you suspected, you can use the corresponding statement for the trace on endomorphism valued forms, which itself follows from a simply standard computation. Your statement then follows from the fact, that your metric gives an isomorphism (I assume your bundles are real) between $E^*$ and $E$ such that the connection induced on $E^*$ by the isomorphism is actually the dual connection.

Of course, you can perform the computation directly, as in your attempt. Write your form locally as a finite sum of forms of the following type: $$e\otimes f\otimes \omega$$ where $e,f$ are locally defined sections in $E$ and $\omega$ is a closed $j$-form. Then $$d^\nabla(e\otimes f\otimes \omega)=(d^\nabla(e\otimes f))\wedge\omega$$ and $$d\, tr_\eta(e\otimes f\otimes \omega)= d(tr_\eta (e,f)\otimes \omega)=d(tr_\eta (e,f))\wedge \omega$$ and the statment follows from the special case $j=0.$

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  • $\begingroup$ If I am not mistaken, there should be additional terms in both your equations, which follow from the Leibnitz rule (and involving the derivatives of $\omega$). $\endgroup$ – Asaf Shachar Oct 24 '16 at 7:42
  • $\begingroup$ Maybe I could have high-lighten it, but I take $\omega$ to be closed, which is possible. $\endgroup$ – Sebastian Oct 24 '16 at 8:12
  • $\begingroup$ Sorry, I missed this. Is there a simple reason for why every form can be expressed locally as a combintation of closed forms? $\endgroup$ – Asaf Shachar Oct 24 '16 at 8:34
  • $\begingroup$ OK, it is obvious since the combination is over $C^{\infty}(M)$, not over $\mathbb{R}$. $\endgroup$ – Asaf Shachar Oct 24 '16 at 8:52

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