The curvature of a tensor product of vector bundles

Question

Let $\nabla_i: \mathcal{A}^0(E_i)\rightarrow\mathcal{A}^1(E_i)$ be linear connections on the complex vector bundles $E_1$, $i=1, 2$, respectively. Then $\nabla=\nabla_1\otimes 1+1\otimes\nabla_2$ is a linear connection on $E_1\otimes E_2$. If $F_{\nabla_i}$ denotes the curvature of $\nabla_i$, prove that $F_{\nabla}=F_{\nabla_1}\otimes 1+1\otimes F_{\nabla_2}$.

My (half-)solution is as follows:

$F_{\nabla}(s_1\otimes s_2)=\nabla(\nabla(s_1\otimes s_2))=\nabla(\nabla_1(s_1)\otimes s_2+s_1\otimes\nabla_2(s_2))$, $s_1\otimes s_2\in\Gamma(E_1\otimes E_2)$. How do you proceed further? Does anyone know?

Answer 1

Here I post a solution to the simple problem I posed, based on Ted Shifrin's hints and remarks.

Proof: For an arbitrary section $s_1\otimes s_2\in\Gamma(E_1\otimes E_2)$ we have:

$F_{\nabla}(s_1\otimes s_2)=\nabla(\nabla(s_1\otimes s_2))=\nabla(\nabla_1(s_1)\otimes s_2+s_1\otimes\nabla_2(s_2))$.

In the following we compute $\nabla(\nabla_1(s_1)\otimes s_2)$ and $\nabla(s_1\otimes\nabla_2(s_2))$ separately.

We pay attention to the fact that $\nabla_1(s_1)\otimes s_2$ and $s_1\otimes\nabla_2(s_2)$ are no sections of $E_1\otimes E_2$ but 1-forms with values in the tensor bundle $E_1\otimes E_2$ so we cannot use the decomposition $\nabla=\nabla_1\otimes 1+1\otimes\nabla_2$ on them directly.

Instead we use the fact that we can express $\nabla_1(s_1)\in\mathcal{A}^1(X, E_1)$ as a linear combination of elements $\alpha\otimes t$, $\alpha\in\mathcal{A}^1(X)$, $t\in\Gamma(E_1)$ and $\nabla_2(s_2)$ as a linear combination of elements $s\otimes\beta$, $s\in\Gamma(E_2)$, $\beta\in\mathcal{A}^1(X)$, respectively.

We compute then:

$\nabla(\nabla_1(s_1)\otimes s_2)=\sum_i(\nabla(\alpha_i\otimes t_i\otimes s_2))=\sum_i(d\alpha_i\otimes t_i\otimes s_2-\alpha_i\wedge \nabla(t_i\otimes s_2))=\sum_i (d\alpha_i\otimes t_i\otimes s_2-\alpha_i\wedge \nabla_1t_i\otimes s_2 + \alpha_i \otimes t_i\wedge\nabla_2s_2)=\sum_i ((d\alpha_i\otimes t_i-\alpha_i\wedge\nabla_1t_i)\otimes s_2-(\alpha_i\otimes t_i)\wedge \nabla_2(s_2))=\sum_i(\nabla_1(\alpha_i\otimes t_i)\otimes s_2-(\alpha_i\otimes t_i)\wedge\nabla_2(s_2))=\nabla_1(\nabla_1(s_1))\otimes s_2-\nabla_1(s_1)\wedge\nabla_2(s_2)=F_{\nabla_1}(s_1)\otimes s_2-\nabla_1(s_1)\wedge\nabla_2(s_2)~~~~(1)$

where we used the natural extension of $\nabla: \mathcal{A}^k(X, E_1\otimes E_2)\rightarrow\mathcal{A}^{k+1}(X, E_1\otimes E_2)$ given by

$\nabla(\gamma\otimes u)=d\gamma\otimes u+(-1)^{|\gamma|}\gamma\wedge\nabla u$.

Analogously, repeating the same steps as above one gets

$\nabla(s_1\otimes\nabla_2(s_2))=s_1\otimes F_{\nabla_2}(s_2)+\nabla_1(s_1)\wedge\nabla_2(s_2)~~~~(2)$.

Adding (1) and (2) one gets the desired formula. Q.E.D.

Answer 2

Keep going. You'll get four terms. The "cross-terms" $\nabla_1s_1\otimes\nabla_2s_2$ cancel. Why?

EDIT: You get an induced connection on $E\otimes T^*M$ by setting $\nabla'(s\otimes \omega)=\nabla s\wedge\omega +s\otimes d\omega$. Note this is a $2$-form with values in $E$ and the usual rules for passing exterior derivative across $1$-form apply.

Answer 3

Despite that this question has been closed quite a long time ago, I would still like to share an alternative proof which is essentially a computation using local frames.

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Personally I think it would be great for us to be clear what does the term $(\nabla_1\otimes 1)(s_1\otimes s_1)$ mean before we begin the computation. Of course this is just a shorthand of writing $(\nabla_1s_1)\otimes s_2$, but note that $\nabla_1s_1$ is an $E_1$-valued 1-form. Such form can be written as a sum of simple tensors like $\omega\otimes s$, where $\omega\in\mathcal{A}^1(X)$ and $s\in\Gamma(E_1)$. Then for any $t\in\Gamma(E_2)$, the meaning of $(\omega\otimes s)\otimes t$ should be understood as $\omega\otimes(s\otimes t)$; note that \begin{align} \omega\otimes\underbrace{(s\otimes t)}_{\in\Gamma(E_1\otimes E_2)}\in \mathcal{A}^1(X)\otimes\Gamma(E_1\otimes E_2)\simeq\mathcal{A}^1(X,E_1\otimes E_2) \end{align} in indeed an $E_1\otimes E_2$-valued 1-form on $X$. Similar for the term $(1\otimes\nabla_1)(s_1\otimes s_2)$, $(F_{\nabla_1}\otimes 1)(s_1\otimes s_2)$ and $(1\otimes F_{\nabla_2})(s_1\otimes s_2)$.

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Now let $(e_i)$ and $(\epsilon_j)$ be smooth local frames for $E_1,E_2$ respectively. Let $\omega_i^j$ and $\theta_i^j$ be the connection 1-forms of $\nabla_1$ and $\nabla_2$ respectively. Then \begin{align} \nabla(e_i\otimes\epsilon_j) &=\nabla_1e_i\otimes\epsilon_j+e_i\otimes\nabla_2\epsilon_j \\ %%% &=(\omega_i^p\otimes e_p)\otimes\epsilon_j+e_i\otimes(\theta_j^q\otimes\epsilon_q) \\ %%% &=\omega_i^p\otimes(e_p\otimes\epsilon_j)+\theta_j^q\otimes(e_i\otimes\epsilon_q) \end{align} where the last step follows from our discussion above.

Now let $\Omega_i^j$ and $\Theta_i^j$ be the curvature 2-forms of $\nabla_1$ and $\nabla_2$ respectively. The curvature 2-forms are related to the connection 1-forms by the Cartan's structure equation \begin{align} \Omega_i^j=d\omega_i^j-\omega_i^k\wedge\omega_k^j,\qquad \Theta_i^j=d\theta_i^j-\theta_i^k\wedge\theta_k^j \end{align} (Warning: Different conventions on the indices are assumed in different texts, so the minus sign may be a plus sign in these texts. Nevertheless, the choice of convention should not affect the result of computations.) Then we have \begin{align} F_{\nabla}(e_i\otimes\epsilon_j) &=\nabla\left[\omega_i^p\otimes(e_p\otimes\epsilon_j) +\theta_j^q\otimes(e_i\otimes\epsilon_q)\right] \\ %%% &=d\omega_i^p\otimes(e_p\otimes\epsilon_j) +(-1)^1\underbrace{\omega_i^p\wedge\nabla(e_p\otimes\epsilon_j)}_{=I} +d\theta_j^q\otimes(e_i\otimes\epsilon_q) +(-1)^1\underbrace{\theta_j^q\wedge\nabla(e_i\otimes\epsilon_q)}_{=II} \end{align} where the second step follows by the product rule, and the extra minus sign is due to the fact that $\omega_i^j,\theta_i^j$ are ($E_1$-, $E_2$-valued) 1-forms. Using the definition of $\nabla$, it can be computed that \begin{align} I&=\omega_i^p\wedge\left(\nabla_1e_p\otimes\epsilon_j +e_p\otimes\nabla_2\epsilon_j\right) \\ %%% &=\omega_i^p\wedge\left(\omega_p^r\otimes e_r\otimes\epsilon_j +\theta_j^r\otimes e_p\otimes\epsilon_r\right) \\ %%% &=\left(\omega_i^r\wedge\omega_r^p\right)\otimes\left(e_p\otimes\epsilon_j\right) +\left(\omega_i^p\wedge\theta_j^q\right)\otimes\left(e_p\otimes\epsilon_q\right) \end{align} The same computation also applies to $II$. As a result, we obtain \begin{align} F_{\nabla}(e_i\otimes\epsilon_j) &=d\omega_i^p\otimes(e_p\otimes\epsilon_j) -\left(\omega_i^r\wedge\omega_r^p\right)\otimes\left(e_p\otimes\epsilon_j\right) -\left(\omega_i^p\wedge\theta_j^q\right)\otimes\left(e_p\otimes\epsilon_q\right) \\ &\qquad+d\theta_j^q\otimes(e_i\otimes\epsilon_q) -\left(\theta_j^q\wedge\omega_i^p\right)\otimes\left(e_p\otimes\epsilon_q\right) -\left(\theta_j^r\wedge\theta_r^q\right)\otimes\left(e_i\otimes\epsilon_q\right) \\ %%% &=\left(d\omega_i^p-\omega_i^r\wedge\omega_r^p\right)\otimes(e_p\otimes\epsilon_j) \\ &\qquad +\left(d\theta_j^q-\theta_j^r\wedge\theta_r^q\right)\otimes(e_i\otimes\epsilon_q) \\ &\qquad -\left(\omega_i^p\wedge\theta_j^q+\theta_j^q\wedge\omega_i^p\right) \otimes\left(e_p\otimes\epsilon_q\right) \end{align} The last term clearly vanishes, and the first two terms are simplified by the Cartan's structure equation: \begin{align} F_{\nabla}(e_i\otimes\epsilon_j) &=\Omega_i^p\otimes(e_p\otimes\epsilon_j) +\Theta_j^q\otimes(e_i\otimes\epsilon_q) \\ %%% &=(\Omega_i^p\otimes e_p)\otimes\epsilon_j +e_i\otimes(\Theta_j^q\otimes\epsilon_q) \\ %%% &=F_{\nabla_1}(e_i)\otimes\epsilon_j+e_i\otimes F_{\nabla_2}(\epsilon_j) \\ %%% &=(F_{\nabla_1}\otimes 1+1\otimes F_{\nabla_2})(e_i\otimes\epsilon_j) \end{align} where the 2nd step follows by our discussion in the beginning while the 3rd step follows from the fact that the curvature $F_{\nabla_i}$ is locally represented by the curvature 2-forms.

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Finally, if one views the curvature $F_{\nabla}$ as a map $\mathcal{A}^0(X;E)\to\mathcal{A}^2(X;E)$, then it is linear over $C^{\infty}(X)$. So the result also holds for arbitrary section in $\Gamma(E_1\otimes E_2)$. This proves the desired.

Answer 4

Here is what I think:

Let $\nabla_{1}=d_{1}+\omega_{1}$, $\nabla_{2}=d_{2}+\omega_{2}$. Then we have $$ \nabla=\nabla_{1}\otimes 1+1\otimes \nabla_{2} $$ Now we have \begin{align} F_{\nabla}(a\otimes b)&=\nabla(\nabla (a\otimes b))\\ &=\nabla(\nabla_{1}a\otimes b+a\otimes \nabla_{2} b)\\ &=\nabla(d_1a\otimes b)+\nabla(\omega_1 a\otimes b)+\nabla(a\otimes \omega_2 b)+\nabla(a\otimes db) \end{align} The first two terms can be further computed by $$ \nabla(d_1a\otimes b)=d_{1}(d_{1}a)\otimes b+(\omega\wedge d_{1}a)\otimes b+d_{1}a\otimes \nabla_{2}b=(\omega\wedge d_{1}a)\otimes b+d_{1}a\otimes \nabla_{2}b $$ as well as $$ \nabla(\omega_1 a\otimes b)=((d_{1}\omega_{1})a-\omega_{1}\wedge d_{1}a)\otimes b+\omega_{1}\wedge \omega_{1}a\otimes b+\omega_{1}a\otimes \nabla_{2}b $$ It is clear that when we sum that the first two terms in the first equation and the first two terms in the second equation now adds up to $$ F_{\nabla_{1}}a\otimes b=F_{\nabla_{1}}(a\otimes b) $$ similarly the second two terms contribute $$ a\otimes F_{\nabla_{2}}b=(1\otimes F_{\nabla_{2}})(a\otimes b) $$

We now have to work with the cross terms. But they are of the form $\nabla_{1}a\otimes \nabla_{2}b+\nabla_{1}a\otimes \nabla_{2}b=0$. So you get the desired identity. However, it is not entirely clear to me why the cross terms would cancel. Maybe I am repeating the same mistake for treating wedge product as tensor product, and there is a minus somewhere.