Intereting Posts

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Ellipses given focus and two points
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The Quaternions and $SO(4)$
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Homomorphism between $A_5$ and $A_6$
On solvable octic trinomials like $x^8-5x-5=0$
Proof that if $\phi \in \mathbb{R}^X$ is continuous, then $\{ x \mid \phi(x) \geq \alpha \}$ is closed.
If $f:\mathbb{R}\to\mathbb{R}$ is a left continuous function can the set of discontinuous points of $f$ have positive Lebesgue measure?
Random matrices, eigenvalue distribution.
definition of strongly convex
Prove that the polynomial $f_n(x)=nx^{n+1}-(n+1)x^n+1$ is divisible by $(x-1)^2$
What is the affine connection, and what is the intuition behind/for affine connection?
Proof $\sum\limits_{k=1}^n \binom{n}{k}(-1)^k \log k = \log \log n + \gamma +\frac{\gamma}{\log n} +O\left(\frac1{\log^2 n}\right)$

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?

- Give an example of a discontinuous bilinear form.
- Diagonalizing symmetric real bilinear form
- Cauchy Schwarz inequality for random vectors
- Should isometries be linear?
- Is every symmetric bilinear form on a Hilbert space a weighted inner product?
- Show that a Bilinear form is Coercive

- Fundamental solution of the Laplacian on the surface of a cylinder
- Proving continuity/smoothness for a special function on a Lie group.
- Given a helix, consider the curve of it's tangent. Express the curvature and torsion of such a curve.
- Pullback distributes over wedge product
- Covariant and partial derivative commute?
- Umbilics on the ellipsoid
- Geodesics on the torus
- What is the intuitive meaning of the scalar curvature R?
- Is it possible to elementarily parametrize a circle without using trigonometric functions?
- uniqueness of the smooth structure on a manifold obtained by gluing

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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