It is a quotient - of the tensor algebra, obtained by taking graded sum over whole numbers $n$ of $n$-fold tensor products - by the ideal generated by elements of the form $a\otimes a$. We write the residue class of $a\otimes b$ in this algebra, as $a\wedge b$ and call it the wedge product.

learn more… | top users | synonyms (1)

3
votes
2answers
116 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
10
votes
3answers
541 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 ...
2
votes
1answer
83 views

Pullback expanded form.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ According to Daniel ...
47
votes
5answers
4k views

Exterior Derivative vs. Covariant Derivative vs. Lie Derivative

In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them ...
7
votes
1answer
166 views

A scalar product in the space of oriented volumes?

Let $L\colon \mathbb{R}^n \to \mathbb{R}^N$ be an injective linear map. By the Cauchy-Binet formula, $\det(L^TL)$ equals the sum of the squares of all minors of $L$ of order $n$: this looks just like ...
7
votes
1answer
164 views

Determinant of the transpose via exterior products

Let $V$ be a finite-dimensional vector space over $F$ and let $\tau:V \to V$ be a linear operator. Here's my definition of the determinant: If $t:U \to U$ is a linear operator and $\dim(U)=n$ then ...
5
votes
0answers
58 views

What about other symmetric functions of the eigenvalues? [duplicate]

Possible Duplicate: Identities for other coefficients of the characteristic polynomial Let $A$ be a matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. Then $\det(A) = \lambda_1 \dots ...
3
votes
1answer
406 views

Understanding of exterior algebra

Consider the following definition from Loring W. Tu's An Introduction to Manifolds: For a finite-dimensional vector space $V$, say of dimension $n$, define ...
2
votes
1answer
31 views

A basis in the $k$-th exterior power of a vector field

Definition: Let $\mathbb R^n$ be the $n$-dimensional real vector space. An exterior $k$-form call any skew-symmetric tensor on $\mathbb R^n$ of rank $k$. Denote the set of exterior $k$-forms by $E^k$. ...
4
votes
1answer
140 views

Are projective modules over exterior algebras of vector spaces necessarily free?

Let $E(V)$ be the exterior algebra of a vector space $V$ (I've also seen this denoted $\Lambda(V)$).Is it true that any projective $E(V)$-module is necessarily free? If it's any easier, is it at least ...
4
votes
1answer
737 views

Is there a formula for the determinant of the wedge product of two matrices?

I was going over the Wikipedia page for exterior products of vector spaces and we can define the determinant as the coefficient of the exterior product of vectors with respect to the standard basis ...
3
votes
1answer
105 views

Transformation rule for a wedge product

Suppose two sets of covectors on a vector space $V, \beta^1,...\beta^k $ and $\gamma^1,...,\gamma^k,$ are related by $$\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$$ where $i=1,...,k$, for a $k\times k$ ...
3
votes
1answer
194 views

What is a more formal name for the wedgey group?

The rule $(v_1,w_1)⋅(v_2,w_2)=(v_1+v_2,w_1+w_2+(v_1∧v_2))$ defines a group structure on the vector space $V⊕(V∧V)$ whenever $V$ is itself a vector space over some field $F$. What is a more common ...
2
votes
1answer
130 views

decomposable elements of $\Lambda^k(V)$

I have conjecture I have problem to prove or disprove. Let's $ w \in \Lambda^k(V)$ is k-vector. $W_w=\{v\in V: v\wedge w = 0 \}$ is k-dimensional vector space if and only if $w$ is decomposable. ...
1
vote
2answers
58 views

If $\phi_i$s are linearly dependent, $\det [\phi_i(v_j)] = 0$ - is the proof legit?

Given $v_1, \ldots, v_k \in V$ and $\phi_1, \ldots, \phi_k \in V^*$. If $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, proof $\det[\phi_i(v_j)] = 0.$ Here $k$ is the dimension of $V$, but ...