# Trace of a bilinear form?

I'm just a beginner of differential geometry, so please forgive me if this is nothing but a silly question or I'm making a critical conceptual mistake.

Let $\mathrm{I\!I}(X, Y)$ be the second fundamental form on an embedded $n$-manifold in the $(n+1)$-dimensional Euclidean space. Then the mean curvature $H$ is defined as the trace of $\mathrm{I\!I}( \cdot, \cdot)$ divided by $n$. But my question arises here. What does the trace of a bilinear form means, especially in the sense of a tensor field on a coordinate chart? Of course, considering $\mathrm{I\!I}(\xi^i g_i, \eta^j g_j) = \mathrm{I\!I}(g_i, g_j) \xi^i \eta_j$ gives us a matrix representation $h_{ij} = \mathrm{I\!I}(g_i, g_j)$ with respect to the contravariant basis. But if we take $H$ as a mere trace of this matrix $(h_{ij})$, we only have a quantity that is not invariant under the change of coordinate, which seems obviously undesirable. I know that I have trouble understanding the nature of $\mathrm{I\!I}( \cdot, \cdot)$, since the correct answer would be $(1/n)h_{i}^{i}$, where $h_{i}^{j} = h_{il}g^{lj}$. Is there any generous soul who can help me out from this conceptual messup?

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You are aware that you can simply compute the mean curvature locally as $H = g^{ij}h_{ij}$, and this is of course invariant? – Glen Wheeler Apr 30 '11 at 20:03

This is a purely local issue. Let $V$ be a finite-dimensional real inner product space with inner product $\langle -, - \rangle$. Then endomorphisms $T : V \to V$ can be naturally identified with bilinear forms on $V$ via the identification $T \mapsto \langle -, T(-) \rangle$. The inverse identification exists thanks to the "Riesz representation theorem" (trivial in this setting). In particular, the trace of a bilinear form can be identified with the trace of the corresponding endomorphism, and so is well-defined up to orthogonal change of coordinates.

Another way of saying this is as follows. You are correct that bilinear forms $V \times V \to \mathbb{R}$ don't have a well-defined notion of trace for $V$ only a real vector space; what has a well-defined notion of trace is an endomorphism $V \to V$, and this is because we can identify endomorphisms with elements of $V \otimes V^{\ast}$, and the dual pairing gives a distinguished map $V \otimes V^{\ast} \to \mathbb{R}$. Because one does not need to make any choices to define this map, it is automatically invariant under change of coordinates.

Bilinear forms, on the other hand, are elements of $V^{\ast} \otimes V^{\ast}$, and no analogue of the dual pairing exists here in general. However, if $V$ is an inner product space, the inner product gives a distinguished isomorphism $V \simeq V^{\ast}$ sending $v \in V$ to $\langle -, v \rangle$ and then the identification above is possible.

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You gave the linear algebra reason. Geometrically, an orthogonal change of (local) coordinates is (by definition) a local isometry, hence it should not change geometric quantities such as curvature (indeed geometrically meaningful quantities should be invariant under local isometries). On the other hand, if you allow non-orthogonal changes you might end up scaling some directions, hence you should also expect change of curvature in general. – t.b. Apr 30 '11 at 17:07
Thanks! This identification clearly yields the 'correct' representation of $\mathrm{I\! I}$ for calculating the trace, namely $(h_{i}^{j})$. – Sangchul Lee Apr 30 '11 at 17:08
@sos440: yes, as long as you compute everything using an orthonormal basis. – Qiaochu Yuan Apr 30 '11 at 18:09