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.

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283 views

Exterior Algebra as quotient

Given a vector space $W$, I understand what the tensor algebra $T(W)$ is, and I understand that the exterior algebra $\bigwedge W$ is defined as $\bigwedge W := T(W)/N$ where $N$ is the two-sided ...
6
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3answers
455 views

Exterior algebra of a vector bundle

Associated to any vector space $V$ is its exterior algebra $\Lambda(V)$ which has the direct sum decomposition $\Lambda(V) = \bigoplus_{i=0}^n\Lambda^i(V)$ where $n = \dim V$. My first interaction ...
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1answer
206 views

Wedge product of forms

Let $\alpha = y^2 dx + dy \in \Omega^1(R^2)$, $\beta = xy dx \wedge dy \in \Omega^2(R^2)$. Is $\alpha \wedge \beta = 0$?
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1answer
59 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 ...
6
votes
2answers
219 views

$\alpha\wedge\beta = 0$ for all $\beta$ implies $\alpha = 0$ without using the Hodge dual

Let $\alpha$ be a differential $k$-form on an orientable smooth $n$-dimensional manifold. If $\alpha\wedge\beta = 0$ for every differential $(n - k)$-form $\beta$, then $\alpha = 0$ because we can ...
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0answers
144 views

Multiplication in exterior algebra

Take $V = K^{n}$. Let $\omega$ be a non-zero element of $\bigoplus_{k=1}^n \bigwedge^k V$, where we have excluded the summand $\bigwedge^0 V = K$. (1) Prove that there exists an $m > 1$ for which ...
6
votes
2answers
327 views

algebraic manipulation of differential form

Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ $(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes ...
2
votes
1answer
284 views

Wedge product with a non-degenerate form

Let $\alpha$ be a non-degenerate form in $\Lambda^k(V)$ for some vector space $V$, $\dim V = n$. (Here non-degenerate means that if $x\in V$ is nonzero, then $(y_1 , ... , y_{k-1}) \mapsto \alpha(x , ...
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1answer
60 views

How can I show that $\det(v_1,v_2,\ldots,v_n)=dx_1\, dx_2\cdots dx_n(v_1,v_2,\ldots,v_n)$?

I wanted to use the definition of a wedge product which says $λ_1λ_2\cdots λ_k(v_1,v_2,\ldots,v_k)=\det(λ_i(v_j))$ with $1<i,j<k$ but I'm not sure if that even can work
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187 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 ...
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0answers
150 views

Explicit computations of tensor and wedge product

Let $f\colon K^3\to K^3$ be a map in Jordan canonical form having matrix $$ f=\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0& -1 \end{bmatrix} $$ What is $f\otimes f$? What ...
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6answers
7k 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 ...
2
votes
1answer
317 views

Taking the exterior derivative of a 0-form

I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative. The $f_i$ are ...
4
votes
3answers
167 views

elementary question regarding differential forms

Is it possible to give a high level explanation why changing the order of differentials will give rise to a minus sign ? I.e. why do we have $$ dx\,dt = - dt\,dx $$ (I am going to take a course on ...
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votes
1answer
102 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
106 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
419 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 ...
5
votes
0answers
67 views

Inner product of $p$-forms [duplicate]

Possible Duplicate: Extension of Riemannian Metric to Higher Forms I have no problems with understanding the inner product of 1-forms on a Riemannian manifold. We have a metric tensor, it's ...
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votes
1answer
171 views

$\wedge^k(V)^* \cong \mathrm{Alt}^k(V)$

Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$. I ...
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0answers
65 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 ...
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votes
1answer
696 views

Interpreting the determinant of matrices of dot products

In the Euclidean space $\mathbb{R}^n$ consider two (ordered) sets of vectors $a_1 \ldots a_k$ and $b_1 \ldots b_k$ with $k \le n$. Question What is the geometrical interpretation of ...
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0answers
163 views

Calulation of pullback of form

If $M$ is $2n+1$ dimensional manifold, and $M'= M\times \mathbb R$ Let $x_1,y_1,... x_n, y_n,t', t$ be coordiante of $M'$. With $t$ for coordinate for $\mathbb R$. Let $$ \omega= \sum_{i=1}^n ...
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1answer
880 views

wedge product of differential form

If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ ...
2
votes
2answers
184 views

Second exterior power of a complex vector space

Suppose we are given a $k$-dimensional complex vector space. Consider the second exterior power of $V$, that is $\Lambda^2V$. Denote $X=\{v_1\wedge v_2\colon v_1,v_2\in V\}$. Now I have two ...
6
votes
1answer
118 views

Wedge product of Hochschild Cohomology classes in characteristic 2

Let $A$ be a smooth commutative $k$-algebra, for $k$ a commutative ring. By the Hochschild-Kostant-Rosenberg theorem, we have that $HH^*_k(A)\cong \Lambda^* \mathrm{Der}_k(A,A)$, where ...
7
votes
1answer
189 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 ...
3
votes
1answer
111 views

Is $\bigwedge(V)$ self-injective?

For a vector space $V$, is the grassman algebra $\bigwedge(V)$ always an injective module over itself? Is there a proof, or even just a brief explanation?
4
votes
1answer
158 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 ...
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vote
1answer
213 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 ...
3
votes
1answer
142 views

What is the difference between $\ker( L \bigwedge L \overset{[-,-]}{\rightarrow} L )$ and $\langle a \wedge b \big| [a,b]=0\rangle$?

Let $L$ be a finite dimensional Lie algebra. We view the Lie bracket as a linear map on the exterior square: $$\pi:L \bigwedge L \rightarrow L$$ Define $$\bigwedge L := \langle a \wedge b \big| ...
2
votes
1answer
272 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 ...
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2answers
271 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 ...
2
votes
0answers
63 views

Determining explicitly the action on the exterior products of a vector space

Let $V$ be a 2-dimensional complex vector space with basis $e_1,e_2$. Consider the endomorphism $f:V\to V$ given by $f(e_1) = e_2$ and $f(e_2) = -e_1$ with matrix $$ \left( \begin{matrix} 0 & -1 ...
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56 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. ...
21
votes
3answers
511 views

What's the connection between derivatives and boundaries?

The (second) fundamental theorem of calculus says that $$\int_a^b f'(x) dx = f(b) - f(a)$$ which can also be stated, if one knows enough about what's coming next, as: The integral of the ...
5
votes
1answer
942 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?
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1answer
579 views

“And” symbol? Wedge product in a surface integral? — Is this a typo, or did I miss an important lecture?

This is the question I got on my final assignment (Calculus III): Evaluate the surface integral \begin{equation} \int \int_S xy \; \; dy\wedge dz - yz \; \; dz\wedge dx + xz \; \; dx\wedge ...
4
votes
1answer
221 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 ...
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0answers
112 views

Given a surjective linear mapping of free modules how do you show the corresponding matrix has an invertible minor?

The following post can is related to part c) of this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=124137 and boils down to some issues I am having with use of wedge ...
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347 views

Using the notation of wedge product to solve a linear system of equations

I am trying to solve a problem that seems like a standard idea from linear algebra but with a the notion of wedge product and exterior algebra added it gets more complicated for someone who isn't ...
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1answer
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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 ...
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votes
1answer
156 views

Associativity of Moyal-like products

The Moyal product of two smooth functions $f,g$ on $\mathbb R^{2n}$ can be defined as $$ f\star g = \exp\left(-\omega^{ij} \frac{\partial}{\partial y^i} \frac{\partial}{\partial z^j}\right) f(y)g(z) ...
3
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1answer
147 views

Parametrizing the total orders on a real vector space

The following question appears trivial, but it's outside my limited experience so I'd appreciate a little feedback. A total order on a real vector space $V$ is a total ordering on its vectors which ...
6
votes
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299 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 $p+q \le n$.
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exterior product definition

i have question from vector mathematics,i know that if there is given two vector, for instance $a=\{a_1,a_2,a_3\}$,$b={b_1,b_2,b_3}$; then so called exterior product is determined as $a\wedge ...
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1answer
543 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 ...
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376 views

Understanding of graded algebra

I am recently learning from Loring W. Tu's An Introduction to Manifolds the concept graded algebra, which is used for introducing exterior algebra. I don't understand the following definition: An ...
12
votes
2answers
867 views

Grassmann numbers as eigenvalues of nilpotent operators?

The following question is motivated by the construction of the fermionic path/field integral, as done for example in Altland & Simons "Condensed Matter Field Theory". Consider the vector space ...
12
votes
3answers
673 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
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0answers
143 views

Exterior algebras and radicals

So without giving excessive background, I was working on baby Spivak & got to the stuff about exterior algebras, & now I'm going through the section on tensor algebras in my copy of ...