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I think there is an importan point that has been overlooked in the above answers: The exterior derivative is the only linear natural operator in the list. This is explained with several variations the book by Kolar, Michor and Slovak cited in Yuri Viatkin's answer. The Lie derivative is also natural under general diffeomorphisms but only as a bilinear ...

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It should be stated in the problem but I think you are right about $v_i = \pi (u_i)$. Hint: observe that $\frac{1}{2}((u_1 + u_2) \otimes (u_1 + u_2))= \frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes u_1) + \frac{1}{2}(u_1 \otimes u_1 + u_2 \otimes u_2)$ is projected by $\pi$ to $0$ on one hand (projection of LHS) and to $\frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes ... 1 As darij grinberg mentions, the exterior algebra is the free graded-commutative algebra on a vector space. It's not the free algebra or the free graded algebra. This doesn't rule out the possibility of "unexpected" right adjoints. Your dimension-counting argument doesn't work, because in order for$\operatorname{dim}(A/B) = \operatorname{dim} A - ...

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An $R$-algebra structure on $M\otimes M$ would looks like a map $(M\otimes M) \otimes (M \otimes M) \to M\otimes M$. The natural structure map you are equipped with is merely a bilinear map $M \times M \to M\otimes M$, which does not have the right domain to tell you how to define how to multiply two basic tensors $m_1 \otimes m_2$ and $m_3\otimes m_4$. ...

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Like GFR, I also have no interest in an extended discussion. This post is meant to be an extended comment for the sake of adding perspective. Personally, I like to define the exterior derivative $d$ in a coordinate-free, invariant way: Fact/Def: Let $M$ be a smooth manifold (possibly with boundary). Then there are unique operators $d \colon ... 1 I am not going to get engaged in a full rebuttal, for no other reason that it would take too much time/efforts, but a few points: The symbol$\nabla$is usually used for covariant derivatives, so$d= \partial\wedge$would be better. In fact I have seen this notation used somewhere, but cannot remember where. So yes, you could use this notation, personally ... 0 Here is my partial answer or conjecture. Let us look for a general notion of "symmetrization" (bad term?) of a function from a set to an Abelian semigroup in the following sense. Given a finite group$G$with a subgroup$H$, a set$X$with a right$G$-action (a$G$-set), an Abelian semigroup$M$with a right$G\$-action by automorphisms (an Abelian ...

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