Tag Info

Hot answers tagged

2

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 ...


2

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 ...


1

The answer to this question turned out to be rather uninteresting. There was an error on one of the Wikipedia articles which I had consulted, and this lead to my confusion. By the way, having done a bit more reading, I think that using the Pl├╝cker relations the way that I did, i.e. to decide when an element $\alpha \in \Lambda^2(X)$, $X$ a ...


1

As Darij remarked in his comment, if $R$ is not a ${\mathbb Q}$-algebra one cannot argue that $\bigwedge^n P$ and ${\mathfrak S}^n P$ are summands of $P^{\otimes n}$, thereby reducing the statement to showing that projectivity and flatness are stable under taking tensor powers. Instead, one might argue as follows: Freeness: If $F$ is a free $R$-module on ...


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 - ...



Only top voted, non community-wiki answers of a minimum length are eligible