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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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 ...
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228 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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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 ...
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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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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. ...
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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 ...
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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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Grassmann Algebras

The Grassmann algebra $G$ is the algebra over a field $\mathbb{F}$ generated by the variables $e_i$ such that $e_i^2=0$ and $e_i e_j = - e_j e_i$. I'm looking for some references on algebras $G ...
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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 ...
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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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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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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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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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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) ...
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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 ...
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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 ...
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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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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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Characterization of rank in an exterior algebra

The wikipedia page on exterior algebras makes the following reasonable sounding statement (I paraphrase): Let $V$ be a complex vector space and consider the second exterior power $\bigwedge^2 V$. By ...
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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 ...
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Morphism of Exterior Algebras

Let $k$ be a field, let $V$ and $W$ be $k$-vector spaces of dimensions $n$ and $m$ respectively, and let $f:V\to W$ be a $k$-linear transformation. Let $\Lambda(V)$ and $\Lambda(W)$ denote the ...
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Wedge Product, A Novel Interpretation or Just Plain Wrong?

I have read (I think) all of the previous threads on this website (and many others) on this topic & unfortunately have not found an answer to my question. Due to the fact that I am only beginning ...
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Wedge product and cross product - any difference?

I'm taking a course in differential geometry, and have here been introduced to the wedge product of to vectors defined (in Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo) ...