# Tag Info

## Hot answers tagged differential-geometry

12

Your $M$ is not compact, and $\omega$ is not compactly supported. So you have not contradicted Stokes's theorem.

10

Congratulation! You found something which leads to idea of closed but not exact forms and cohomology groups. To see the problem take $M= \overline{B^2(1)} \setminus B^2(\epsilon)$. Than $\partial M = S^1(1) - S^1(\epsilon)$. Now the Stokes theorems says: $$\int_{S^1(1)} \omega - \int_{S^1(\epsilon)} \omega = \int_M d\omega$$ You might think that integral ...

4

For (nice) space curves with a Frenet frame, by convention, curvature $\kappa$ is always positive. There are two ways in which "signed curvature" is used to refer to curves. (1) For regular plane curves, we can decide that the unit tangent vector $T$ and principal normal vector $N$ will always make a right-handed basis. Then $T'(s)=\kappa(s)N(s)$, and ...

3

Ok, I undestood what is going on here. It actually a triviality, but a very confusing one. Everything is clear if one have in mind that $\tilde{P}$ is actually a multilinear function. Recall that the differential of a linear transformation is actually the same linear transformation. For example $$T:\mathbb{R^n}\rightarrow \mathbb{R}$$ then the differential ...

3

No curve between two points on a Riemannian manifold ever maximizes arc length. Given any curve $\gamma$ from $p$ to $q$, you can always find a longer curve, e.g. by leaving $\gamma$ for a little while and taking a detour, and then coming back and rejoining $\gamma$ where you left off. The dashed curve in your drawing is a saddle point for the distance ...

3

It suffices to take $\alpha = f\, dx^{i_1}\wedge\dots\wedge dx^{i_p}$ and $\beta=g\,dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Then $\alpha\wedge\beta = fg\, dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $d\alpha\wedge\beta$ and $\alpha\wedge ... 3 When changing the basis with respect to which a bilinear form is represented, we have$$A'=P^tAP,$$where$A$and$A'$are the representing matrices and$P$is the transition matrix. When one of the matrices represents$g$with respect to an orthonormal basis, we have$$A'=P^tIP,$$ and thus$$\det A'=\det P^t\cdot\det P=(\det P)^2.$$The last sentence in the ... 2 The vector fields$Y_i$do necessarily span an integrable distribution$E \subseteq T(U \times \overline{U})$.$E$is necessarily a smooth subbundle of$T(U \times \overline{U})$, so it suffices to show$E$is closed under the Lie bracket. Note that$[X_i, \overline{X}_j] = 0$for every$i$,$j$because$X_i$only differentiates with respect to the first$n$... 2 First of all, note that in the formula you give us, $$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n},$$ there is a summation involved! In fact, there are$n-p$sums, since the indices are being contracted. I won't use Einstein's convention in ... 2 Sorry to resurrect, but we leave a$($detailed$)$proof here that$TM$has the structure of an oriented$2n$-manifold, even if the$n$-manifold$M$is non-orientable. Let$\{(U_\alpha, \phi_\alpha)\}_{\alpha \in A}$be a smooth atlas of$M$, and let$V_\alpha = \phi_\alpha(U_\alpha) \subset \mathbb{R}^n$. Then$(\phi_\alpha)_*: TU_\alpha \to TV_\alpha = ...

2


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