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Given a manifold $M$ with hermitian metric $g$, we can extend this metric to the cotangent space $T^*M$ and to all vector bundles $\Lambda^{p,q}T^*M$ via the following (See Griffiths and Harris p. 80). Let $$ \omega = \frac{i}{2} \sum \phi_j \wedge \bar{\phi_j} $$ be the associated (1,1) form to the metric, and $\{\phi_1, \ldots, \phi_n\}$ a unitary coframe. Then define a metric on $\Lambda^{p,q}T^*M$ given by $\{\phi_I\wedge \bar{\phi_J}\}$ is orthogonal where $\vert I\vert = p$ and $\vert J\vert = q$, and by $\|\phi_I\wedge\bar{\phi_J}\|^2 = 2^{p+q}$.

To extend this globally, we define the global inner product to be $$ (\alpha, \beta) = \int_M (\alpha(z), \beta(z))\text{dVol}. $$

My question is, how do we extend this to matrix-valued forms? In particular, how can I define a norm of the curvature of a given connection? That is, how do I get a metric on $\Lambda^2 T^*M \otimes \text{End}(E)$? (E is a complex vector bundle).

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  • $\begingroup$ Is $E$ an arbitrary (complex?) vector bundle? $\endgroup$ – user309475 Jul 2 '17 at 21:52
  • $\begingroup$ Yes it is. Sorry. $\endgroup$ – user46348 Jul 2 '17 at 21:55
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I assume that you are given a bundle metric on your $E$.

Given two vector bundles $V$ and $E$ over a manifold $M$, equipped with bundle metrics $\langle.,.\rangle_V$ and $\langle.,.\rangle_E$, respectively, there is an induced tensor product metric on the vector bundle $V\otimes E$, which is defined on elementary tensors by

$\langle v_1\otimes e_1,v_2\otimes e_2\rangle_{V\otimes E}:=\langle v_1,v_2\rangle_V\langle e_1,e_2\rangle_E$

where $v_i\in V$, $e_i\in E$.

Matrix valued forms are a special case of this.

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