Intereting Posts

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Inequality of numerical integration $\int _0^\infty x^{-x}\,dx$.
Probability mean is less than 5 given that poisson distribution states it is 6
Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions?
Finding the basis of $\mathfrak{so}(2,2)$ (Lie-Algebra of $SO(2,2)$)
How to Evaluate $ \int \! \frac{dx}{1+2\cos x} $ ?
Proof by induction that $\sum\limits_{k=1}^n \frac{1}{3^k}$ converges to $\frac{1}{2}$
What should an amateur do with a proof of an open problem?
proving $ (A \rightarrow B \vee C )\rightarrow((A\rightarrow B) \vee (A\rightarrow C))$
Prove that $\mathbb{|Q| = |Q\times Q|}$
Green's first and second identities
Understanding the Definition of a Differential Form of Degree $k$
Tricky inequality no avail to AM-GM
Creating unusual probabilities with a single dice, using the minimal number of expected rolls
How to prove tr AB = tr BA?

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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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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