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Hot answers tagged exterior-algebra

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 Plücker map consists of all (projectivized) decomposable $k$-vectors, which is in general a very thin subset of $\Bbb P(\Lambda^k V)$.

3

Other that simplifying into $\bigwedge R=R\oplus R$, you are completely right. You can also go for a different approach, by defining the degree$-n$ terms as follows. Namely by $\bigwedge^nM=M^{\otimes n}/N$, where $N$ is the submodule generated by the dublicate terms. Then we have $\bigwedge M=\bigoplus_n \bigwedge^n M$. From this definition, it is ...

3

You asked "If A wedges B and A wedges C ...". Usually, a statement of the form "If X and Y" only makes sense if X and Y are things that can be true or false, hence our confusion. However, if we interpret it as "If $A \wedge B = A \wedge C$, does $B = C$?" - in other words, if the exterior vector products are equal are the vectors themselves equal? The ...

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

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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