Tagged Questions

2
votes
1answer
48 views

In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes?

I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y ...
3
votes
1answer
66 views

Elements of $\wedge^2V$ expressible in the form $v_1\wedge v_2$

If $V$ is a complex vector space, then an element $w\in \wedge^2V$ is of the form $v_1\wedge v_2$ for some $v_1,v_2\in V$ iff $w\wedge w=0$ in $\wedge^4V$. Could anybody give some intuition/show why ...
1
vote
1answer
20 views

2-form associated with a skew map

Given a two-form $\omega\in \Lambda^2V$ for some (say finite dimensional) vector space $V$ we may associate with $\omega$ a skew map $f_{\omega}:V\rightarrow V^*$ given by $X\mapsto \iota_X\omega$, ...
2
votes
1answer
54 views

Is there a smallest sub-Grassmann algebra containing a given vector in Grassmann algebra?

Let $V$ be a $d$-dimensional $\mathbb C$-vector space and the Grassmann algebra $$\mathcal G (V):=\bigoplus_{n=1}^d V^{\wedge n}$$ where $\wedge$ denotes the antisymmetric tensor product. I was ...
5
votes
1answer
64 views

general trace relation

Let $V$ be vector space $\dim V=N$, and $A\in End(V)$. Denote $$ \wedge^k A^m(\mathbf{v}_1\wedge\dots\wedge\mathbf{v}_k)=\sum_{s_1,\dots,s_k=0,1,\sum_j s_j=m} A^{s_1}\mathbf{v}_1\wedge\dots\wedge ...
4
votes
1answer
55 views

Trace of the multiplication operator

Let $V$ be vector space, $\dim V=N$. Define the multiplication operator $L_{\mathbf{b}}$ as $L_{\mathbf{b}}:\omega\to \mathbf{b}\wedge\omega$, where $\omega\in\wedge V$ ($\wedge V$ is the entire ...
3
votes
1answer
178 views

Laplace expansion

This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix ...
0
votes
1answer
87 views

Exterior Algebra of Self-Direct Sum

Suppose $V$ and $W$ are vector spaces and $\bigwedge V$ and $\bigwedge W$ their exterior algebras. Then it is known that $\bigwedge (V \oplus W) \simeq \bigwedge V \otimes \bigwedge W$. Now my ...
2
votes
1answer
111 views

Basis of the symmetric algebra $S(M)$ given $R$-module basis of $M$ using the diamond lemma?

Over the past week, I read this secret blogging seminar post concerning the diamond lemma, which got me to reading about Bergman's paper on the diamond lemma. Now suppose you have a free $A$-module ...
11
votes
2answers
239 views

Why is it that $\det(\phi-x\text{id})=\sum_{i=0}^n (-1)^ic_ix^i$?

I'm trying to understand a certain formula for the determinant in a more general setting. Say you have a free module $M$ of rank $n$ over a (commutative) ring $R$. Let ...
1
vote
0answers
44 views

Alternating forms tangential to a subspace.

Let $V$ be a finite-dimensional vector space with euclidean product, and let $U$ be a subspace. Now let $P$ be the projection of $V$ onto $U$, and let $\omega$ be any alternating multilinear $k$-form. ...
5
votes
1answer
337 views

Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?
4
votes
1answer
190 views

On Chevalley's linear identification of the Clifford algebra $C(\mathbf p)$ with the exterior algebra $\wedge \mathbf p$

In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following: "...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto ...
5
votes
2answers
213 views

Decomposition of product of exterior products

Suppose $V$ is a $n$-dimensional vector space. What is the kernel of $$\bigwedge^p V \otimes \bigwedge^q V\longrightarrow \bigwedge^{p+q} V$$ here $p+q \le n$.
9
votes
3answers
369 views

Signs in the natural map $\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$

Let $V$ be a finite-dimensional vector space over a field $\Bbbk$. Let $V^*$ denote its dual. I strongly suspect that there is a natural map $$\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$$ that ...