Take $M$ a Riemannian manifold and $\Lambda^1$ its space of one forms. The LCC (Levi-Civita connection) $\nabla:\Lambda^1 \to \Lambda^1 \otimes \Lambda^1$ is well known to satisfy the identity $m \circ \nabla = d$ for $m: \Lambda^1 \otimes \Lambda^1 \to \Lambda^1$ the usual exterior algebra multiplication and $d$ the exterior derivative. The LCC can be extended to higher forms in the standard way $\nabla:\Lambda^k \to \Lambda^k \otimes \Lambda^1$. But does this $\nabla$ still satisfy the obvious higher identity $m \circ \nabla = d$?
$\begingroup$
$\endgroup$
1
asked Sep 30, 2023 at 12:50
-
1$\begingroup$ do you mean $m: \Lambda^1 \otimes \Lambda^1 \to \Lambda^2$? $\endgroup$– Ali TaghaviCommented Oct 8, 2023 at 8:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Both $m∘∇$ and $d$ are natural operations from $k$-forms to $(k+1)$-forms.
By Palais's theorem, all such operations are proportional to the de Rham differential. That is, $m∘∇=λd$ for some $λ$.
By examining the case of ${\bf R}^n$ with the standard metric, we see that the proportionality constant $λ$ equals 1.
answered Sep 30, 2023 at 15:21
-
3$\begingroup$ It is not clear to me why $m\circ\nabla$ is natural, since it appears to depend on $\nabla$ which is not natural. $\endgroup$ Commented Oct 1, 2023 at 8:40
-
$\begingroup$ what is Palais theorem precisely? $\endgroup$ Commented Oct 8, 2023 at 8:10
-
$\begingroup$ @BenMcKay I do not know what is Palais theorem precisely but it seems that your comment somewhat is also applicable to Palais theorem too(according to $m\circ \nabla=\lambda d$ for a constant $\lambda$ $\endgroup$ Commented Oct 8, 2023 at 8:13
-
$\begingroup$ Palais's theorem appeared in Palais, R., Natural operations on differential forms, Trans. Amer. Math. Soc. 92 (1959), 125–141. It says that the only diffeomorphism invariant operators on differential forms are constant multiples of the exterior derivative and the identity operator. $\endgroup$ Commented Oct 8, 2023 at 9:25