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3

Use the Corollary of this paper: If a compact Riemannian manifold $M$ splits as $M = M_1 \times M_2$, then the identity component of the isometry group splits as $I_0 (M) = I_0(M_1) \times I_0(M_2)$. (This is authors' stated Corollary, but from their main theorem it looks like we can weaken compactness to your requirement that $M$ has no Euclidean ...

1

For compact manifolds the boundary is usually assumed to be empty unless stated otherwise. This should answer both questions. (As you figured out yourself the statement is not true otherwise). (Note: a (differentiable) manifold, by definition, is locally homeomorphic (diffeomorphic) to an open neighbourhood of $\mathbb{R}^n$. This is not true for manifolds ...

1

Let $e_1,\cdots e_m$ be local orthonormal basis on $M$ and $e_{m+1} , \cdots e_s$ be a local orthonormal frame normal to $M$. Let $\overline \nabla$ and $\nabla$ be the covariant derivative on $S$ and $M$ respectively. \begin{split} \Delta^S f &= \sum_{k=1}^s \overline\nabla^2 f(e_k, e_k) \\ &= \sum_{k=1}^m \overline\nabla^2 f(e_k,e_k) + ...

1

The tangent space at $I_n$ is the space of antisymmetric matrices defined by $A+A^T=0$. Given $g\in SO(n)$, and let $AS(n)$ the space of antisymmetric matrices, the tangent space at $g$ is $gAS(n)$. That is the image of the tangent space $T_{I_n}SO(n)$ by the left translation defined by $g$. Suppose that $g\in SO(n)$, $A\in AS(n)$, you have ...

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If $g$ is a degenerate metric, then typically there will be lots of connections that are symmetric and compatible with $g$. For example, let $g$ be the constant-coefficient degenerate metric on $\mathbb R^2$ that you mentioned. Then a straightforward computation shows that the standard Euclidean connection $\nabla$ is compatible with $g$. Since it's ...

1

It's always risky to answer "no" to open-ended "is it possible"-type questions. That said, in the case of using the volume formula for a family of regions to deduce surface area (the way the area of a sphere of radius $r$ is the derivative with respect to $r$ of the volume of a ball of radius $r$), the answer is probably "no": Think, for example, of a ...

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