# Tag Info

## New answers tagged exterior-algebra

1

First notice that there exists unique $\gamma:\wedge^{k+1}V^*\rightarrow \wedge^k V^*$ such that $$<\gamma(T),S>=<T,\xi\wedge S>.$$ This is due to the nondegeneracy of inner product $<,>.$ In fact, assume that there are two such $\gamma_1,\gamma_2.$ Fix arbitrary $T\in\wedge^{k+1}V^*.$ Now you get that for every $S\in\wedge^{k}V^*$ ...

1

Usually this comes up in the context of Cayley-Hamilton. See that the left side is exactly the $n-1$ order term in $\lambda$ when one starts expanding $\det ( A - \lambda I)$. But Cayley-Hamilton can be derived not in terms of determinants but traces of linear operators on the exterior algebra. In particular, let $A_{(k)}$ denote the natural analogue of ...

2

Here's a hint on your third thought. How does one differentiate a determinant? One row/ column at a time, holding the others fixed. Let $V = [v_1,v_2,\cdots, v_n]$ where $v_i'$ are columns. Then, look at the matrix $e^{tA}V.$ Can you differentiate this function at $t=0$ in two ways that correspond to the two sides?

0

You can visualize the exterior derivative in terms of a boundary operation. The fact that $d \circ d = 0$ can then be understood as following from the fact that the boundary of a boundary is empty. The idea is difficult to explain without pictures, so here's a short PDF that explains the idea and gives some nice diagrams to help you visualize differential ...

1

Although the map $(x,y)\to *(x\wedge y)$ doesn't generalize to a map $\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ in general, not every cross-product has to be generated as such. In fact, the cross-product on $\mathbb{R}^3$ is unique modulo sign, so you could think of the construction you outline as just an effect of that uniqueness: anything that ...

0

In k-dimensional space this is simply $|\det U|$; for k-parallelotope $\sqrt{\det(U^TU)}$, see a related "Ratio of area formed by transformed and original sides of a parallelogram".

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