The exterior-algebra tag has no wiki summary.
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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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1answer
169 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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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
$$ \omega= \sum_{i=1}^n ...
4
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1answer
279 views
wedge product of differential form
If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$:
$$\omega= d(r^2\alpha)$$
...
2
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2answers
142 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 ...
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1answer
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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 ...
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1answer
107 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 ...
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1answer
82 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?
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1answer
125 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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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 ...
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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
109 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
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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 ...
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0answers
51 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 $$ \left( \begin{matrix} 0 & -1 ...
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Alternating forms tangential to a subspace.
Let $V$ be a finite-dimensional vector space with euclidean product, and let $U$ be a subspace. Now let $P$ be the projection of $V$ onto $U$, and let $\omega$ be any alternating multilinear $k$-form. ...
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3answers
339 views
What's the connection between derivatives and boundaries?
The (second) fundamental theorem of calculus says that
$$\int_a^b f'(x) dx = f(b) - f(a)$$
which can also be stated, if one knows enough about what's coming next, as:
The integral of the ...
5
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1answer
337 views
Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero
How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?
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“And” symbol? Wedge product in a surface integral? — Is this a typo, or did I miss an important lecture?
This is the question I got on my final assignment (Calculus III):
Evaluate the surface integral
\begin{equation}
\int \int_S xy \; \; dy\wedge dz - yz \; \; dz\wedge dx + xz \; \; dx\wedge ...
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1answer
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On Chevalley's linear identification of the Clifford algebra $C(\mathbf p)$ with the exterior algebra $\wedge \mathbf p$
In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following:
"...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto ...
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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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2answers
200 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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1answer
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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
$$
f\star g = \exp\left(-\omega^{ij} \frac{\partial}{\partial y^i} \frac{\partial}{\partial z^j}\right) f(y)g(z) ...
3
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1answer
118 views
Parametrizing the total orders on a real vector space
The following question appears trivial, but it's outside my limited experience so I'd appreciate a little feedback.
A total order on a real vector space $V$ is a total ordering on its vectors which ...
5
votes
2answers
211 views
Decomposition of product of exterior products
Suppose $V$ is a $n$-dimensional vector space.
What is the kernel of
$$\bigwedge^p V \otimes \bigwedge^q V\longrightarrow \bigwedge^{p+q} V$$
here $p+q \le n$.
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2answers
132 views
exterior product definition
i have question from vector mathematics,i know that if there is given two vector, for instance $a=\{a_1,a_2,a_3\}$,$b={b_1,b_2,b_3}$; then so called exterior product is determined as $a\wedge ...
3
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1answer
284 views
Understanding of exterior algebra
Consider the following definition from Loring W. Tu's An Introduction to Manifolds:
For a finite-dimensional vector space $V$, say of dimension $n$, define
...
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3answers
220 views
Understanding of graded algebra
I am recently learning from Loring W. Tu's An Introduction to Manifolds the concept graded algebra, which is used for introducing exterior algebra. I don't understand the following definition:
An ...
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2answers
742 views
Grassmann numbers as eigenvalues of nilpotent operators?
The following question is motivated by the construction of the fermionic path/field integral, as done for example in Altland & Simons "Condensed Matter Field Theory".
Consider the vector space ...
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3answers
367 views
Signs in the natural map $\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$
Let $V$ be a finite-dimensional vector space over a field $\Bbbk$. Let $V^*$ denote its dual. I strongly suspect that there is a natural map
$$\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$$
that ...
2
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0answers
114 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 ...
2
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1answer
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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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1answer
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What are “Super Numbers”?
I'm reading Hyperspace by Michio Kaku and in the chapter on SuperGravity "Super Numbers" are mentioned and are described as a number system where for any super number $a$, $a*a=-a*a$. I was wondering ...
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1answer
106 views
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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1answer
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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 ...
3
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1answer
180 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 ...
3
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0answers
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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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2answers
193 views
How do I evaluate the Clifford product in dimensions greater than 3?
The Clifford product of a pair of vectors $a,b$ is an associative operation defined by
$$ ab = a \cdot b + a \wedge b.$$
In sufficiently low dimensions I am used to being able to define the Clifford ...
4
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1answer
214 views
Can you find a 2-form not written as the wedge of two 1-forms?
I was under the impression that all 2-forms are the wedge (^) of two 1-forms. Is it possible to have a 2-form that you can't write as A^B with A,B 1-forms?
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4answers
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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) ...
