# An identity for the higher form Levi-Civita connection

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 $$m \circ \nabla = d$$?

• do you mean $m: \Lambda^1 \otimes \Lambda^1 \to \Lambda^2$? Commented Oct 8, 2023 at 8:06

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.
• It is not clear to me why $m\circ\nabla$ is natural, since it appears to depend on $\nabla$ which is not natural. Commented Oct 1, 2023 at 8:40
• @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 $\lambda$ Commented Oct 8, 2023 at 8:13