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6

But $G(3,5)$ has dimension $3(5-3)=6$, which is far less than $9$. The key idea you're missing is that the image of the Pl├╝cker map consists of all (projectivized) decomposable $k$-vectors, which is in general a very thin subset of $\Bbb P(\Lambda^k V)$.


2

To add a slightly more concrete perspective to Ted Shifrin's answer, why don't you try the simpler example of $G(2,4)$ instead? There you have 6 minors, giving an embedding into $\mathbf P^5$. But you can check that the 6 minors of a $2 \times 4$ matrix always satisfy a certain degree-2 equation. (I won't write the equation here, but it is easy to look up, ...


1

How is the interior derivative a derivative? I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation. What is the geometric content of Hodge duality? Short version: you're finding the orthogonal complement of whatever ...


1

The definition of $\mathbb P(K^k)$ is the set of one-dimensional subspaces of $K^k$, not of $K^{k+1}$. The $\mathbb P$ operator takes a vector space and returns its set of one-dimensional subspaces. So $\mathbb P(\Lambda^nK^k)$ is just the one-dimensional subspaces of $\Lambda^nK^k$. Either you've been taught the notation wrongly or you've been confused by ...



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