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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27 views

why is $\omega^n \neq 0$ for a nondegenerate 2 form $\omega$. [duplicate]

Let $\omega$ be a nondegenerate alternating $2$-form on an $2n$-dimesional Vectorspace $V$, meaning that for all nonzero $v \in V$ the map $w \mapsto \omega(v,w)$ is not identically zero. Why is the ...
3
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2answers
65 views

Exterior derivative of a form and $d(d\omega)=0$?

We know that in differential geometry, $d^2\omega=0$, where $\omega $ is a form and $d$ is the exterior derivative. However if this form happens to be the exterior derivative of another form $\theta$...
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0answers
19 views

Criterion for Smoothness of a Function Mapping Into an Exterior Power of a Vector Space.

All vector spaces are assumed to be real and finite dimensional. I am trying to show that: Let $V$ be a vector space and $f:\bigoplus^k V\to \bigwedge^k V$ be defined as $$f(v_1, \ldots, v_k)= ...
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2answers
55 views

Are all $k$-vectors in $\mathbb{R}^3$ simple?

The second comment of the accepted answer here, claims that all $k$-vectors in $\mathbb{R}^3$ are simple, that is, they can be written as the wedge product of k vectors. I tried to show this for $2$-...
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1answer
67 views

Determinant defined using multilinear alternating maps, and invertibility of linear endomorphisms

In Jeffrey Lee's differential geometry text on page 353 he defines the determinant in an interesting way using multilinear alternating maps: Suppose $V$ is an $n$-dimensional $k$-vector space over ...
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0answers
18 views

Annihilator for a tensor $T\in\wedge V^{\ast}$ [duplicate]

For $T\in\wedge^{k} V^{\ast}$ the annihilator is set $$an(T)= \{\phi\in V^{\ast}\mid \phi\wedge T=0\}$$ Then I need to prove that $dim(an(T))\leq k$ and is equal iff $T$ is decomposable ($i.e.$, $T=\...
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121 views

Pullback map distributes over wedge product (proof)

To prove that the pullback map distributes with the wedge product is it first best to prove that it distributes over the tensor product and then use the relation $$dx^{\mu_{1}}\wedge\cdots\wedge dx^{\...
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0answers
37 views

Isomorphism between two vector spaces with the wedge products

F is a field not characteristic 2. V and W are F-vector spaces. If A is the vector space with basis the formal symbols v ∧w with v ∈ V and w ∈ W, and B is the subspace spanned by elements of the form ...
4
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0answers
67 views

When does the duality functor commute with the wedge power functor?

When working with modules over a fixed commutative ring, I know that $(M \otimes N)^* \cong M^* \otimes N^*$ provided either $M$ or $N$ is finitely generated projective. Does it follow that $(\...
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1answer
90 views

The adjoint of left exterior multiplication by $\xi$ for hodge star operator

As we know, for $V$ vectoral space and a orientation $\mathcal{O}$ on $V$ and $e_{1},...,e_{n}$, the hodge star operator $\ast:\wedge V^*\rightarrow\wedge V^*$ is defined for $\ast(e_{1}\wedge...\...
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19 views

The space $V^{0}_{p}$ of p times covariant tensors and canonical isomorphisms

I have been studying tensor calculus by myself, but I have found the following claim in my book: The space $V^{0}_{p}=V^{*} \otimes \cdots \otimes V^{*}$ of $p$ times covariant tensors is ...
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0answers
40 views

Does the wedge product of bundle-valued forms induce a universal object?

Given a smooth manifold $M$ and a vector bundle $E$ over $M$, the $C^\infty(M)$-module of $E$-valued $p$-forms on $M$ is defined to be $$\Omega^p(M; E) := \Gamma_M\left( \bigwedge^p T^*M \otimes E \...
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2answers
79 views

Conceptual approach to the formula $\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n)$

I answered this question earlier showing that $$\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n),$$ and while I am happy with my answer, I feel like there should ...
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1answer
261 views

Geometric intuition about the exterior derivative

Let $M$ be a smooth manifold. One $k$-form is a section of the bundle $\bigwedge^k T^\ast M$, that is, if $p\in M$ and $\omega$ is a $k$-form then $\omega_p$ is one $k$-linear alternating real ...
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2answers
1k views

Symmetric and wedge product in algebra and differential geometry

I have been struggling with this issue for a while (and asked a similar question here), but still not found a satisfying answer. The question boils down to: which is the correct identity? $dx \, dy ...
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1answer
59 views

Why should there be a 7-dimensional cross product in the context of exterior algebra?

The three-dimensional cross product can be viewed as the wedge product corresponding to the exterior power $\Lambda^2(\mathbb R^3)$. An explanation that I have come up with for the scarcity of cross ...
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1answer
69 views

Calculating the norm of an exterior product

I am trying to figure out how to calculate this quantity: $$ \frac{\lVert U^{t}_{\mathbf{x_0}}\mathbf{e}_1\wedge U^{t}_{\mathbf{x_0}}\mathbf{e}_2\wedge\ldots\wedge U^{t}_{\mathbf{x_0}}\mathbf{e}_k\...
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0answers
83 views

Tensor algebra becomes an R-algebra. Theorem, Proof, Dummit and Foote

I have the definition of tensor algebra as follows: $T(M) = \bigoplus_{i =1}^{\infty} T^i(M)$, where $M$ is an $R$ module, where $R$ is commutative and contains the element $1$. Finally $T^k(M) = \...
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50 views

some exterior power of a lattice, torsion-free? (about a remark of Serre in Local Fields)

I have a question on a remark of Serre in chapter III §2 of his book Local Fields (p. 49). In this section he wants to define the discriminant of a lattice with respect to a bilinear form. The ...
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1answer
54 views

How do I show this relation between exterior product and the projection of a tensor product

I have troubles understanding this whole problem starting at the definition. We have defined the exterior product as follows: If $\alpha = \pi (a) \in \bigwedge^pV$ and $\beta = \pi(b) \in \bigwedge^...
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0answers
53 views

Alternating bilinear form with wedge product. equality problem

Let $\phi : \textbf{R}^4 \otimes \textbf{R}^4 \rightarrow \textbf{R}$ be an alternating bilinear form. Prove that there exist linear maps $\alpha, \beta :\textbf{R}^4 \rightarrow \textbf{R}$ with $\...
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0answers
42 views

Alternative proof of existence Jordan normal form

Consider the next theorem: Let be $E$ is an $n$-dimensional vector space over $\mathbb R$ and $\alpha$ a 2-vector. Then there is a basis $\sigma_1,\sigma_2,\ldots,\sigma_n$ such that $$\...
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1answer
58 views

Why $M \otimes M$ does not have a ring structure?

I am reading some section about tensor algebras, and I don't have clear the idea on why $M \otimes M$ dont have a ring structure, where $M$ is an $R$-module. R is commutative and $1 \in R$. So far my ...
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votes
2answers
161 views

“Canonical” symmetrization/skew-symmetrization/alternation of multilinear functions

Is there some precise sense in which the "alternation" functor $A$ that maps a multilinear function $f\colon M^d\to N$ to the alternating multilinear function $A(f)\colon M^d\to N$ defined by $$ A(f)...
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1answer
48 views

Rank of an element of the exterior power

Let $X$ be a finite-dimensional vector space and let $\Lambda^p(X)$ be the $p$th exterior power of $X$. My picture of an elementary element $x_1 \wedge \ldots \wedge x_p$ in the exterior power is ...
4
votes
1answer
102 views

Plücker Relation: misunderstanding?

I'm trying to understand exterior algebra better by gaining some "bare hands" understanding of the exterior powers $\Lambda^k(X)$ in more detail when $\dim(X)$ is small. I think so far I understand ...
4
votes
1answer
116 views

Is there an intuitve motivation for the wedge product in differential geometry?

I've recently started studying differential forms and have been looking at differential forms. I'm struggling to understand the motivation for introducing the notion of the wedge product. Does it ...
4
votes
3answers
106 views

Why are $2$-covectors on $\mathbb{R}^3$ decomposable?

How do you show that every $2$-covector $\omega\in\Lambda^2((\mathbb{R}^3)^\ast)$ is decomposable i.e. that $$\omega=u\wedge v$$ for some $u,v\in(\mathbb{R}^3)^\ast$? In general we have $$\omega=\...
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3answers
747 views

Having trouble understanding generalized complex numbers

I'm reading a paper on the generalized complex numbers, but I have trouble in some of its fundamental properties. I have searched wiki but it left me none the wiser. Please see the image below, in ...
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1answer
82 views

What is the exterior algebra?

I am learning differential geometry, and I have difficulty understanding the construction of the exterior algebra of an $n$-dimensional vector space $V$. We have the wedge product $$\wedge:\Lambda^k(...
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votes
2answers
359 views

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

I have a conjecture. I have a problem proving or disproving it. Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$...
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2answers
244 views

What is the anticommutator of the interior product and codifferential (adjoint of exterior derivative)?

What is $\eta=i_X \delta + \delta i_X$ acting on differential forms? Here $\delta$ is the usual Hodge adjoint of the exterior derivative and $i_X$ is contraction of a form with the vector field $X$. ...
9
votes
3answers
281 views

Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$...
4
votes
1answer
119 views

How do we wedge the complex differentials $\mathrm{d}z^i$ and $\mathrm{d}\bar z^{\bar j}$?

By the standard definition of the wedge product as an alternated tensor product, I would think we have $$\tag{1}\mathrm{d}z^i\wedge\mathrm{d}\bar z^{\bar j}=\mathrm{d}z^i\otimes\mathrm{d}\bar z^{\bar ...
0
votes
1answer
69 views

Exterior power of multilinear functions applied to linearly dependent vectors is zero

I'm working on a homework problem, and we are to show that if $T \in \wedge^p V^*$, and $v_1,\ldots,v_p$ are linearly dependent, then $T(v_1,\ldots,v_p) = 0$. What I've got so far: I understand that ...
2
votes
1answer
64 views

How to phrase this identity in differential form language?

If the vector field $\mathbf B$ on $\mathbb{R}^3$ is constant, then the vector field $$ \mathbf A = \frac 1 2 \mathbf B \times \mathbf r $$ satisfies $$ \nabla \times \mathbf A = \mathbf B. $$ This ...
9
votes
1answer
194 views

How to compute Ext over an exterior algebra

I found this question in several places (even on mathoverflow and mathstackexchange), but I never found a satisfying answer. Let $k$ be a field and $V$ a finite dimensional $k$-vectorspace. I ...
0
votes
0answers
53 views

The kernel of an antiderivation on an exterior algebra

This is a simple algebraic question I feel I should be obvious, but maybe isn't. Let $d'\colon V \twoheadrightarrow W$ be a surjective linear map of finite-dimensional vector spaces over a field of ...
0
votes
1answer
109 views

Differential Geometry-Hodge Star

The Hodge star is given by $$*(dx^{i_1}\wedge dx^{i_2}\wedge....\wedge dx_{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2....i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}$$ The question ...
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0answers
141 views

Can one define wedge products using determinants for $n$-forms?

I was talking to Ted Shifrin in math chat yesterday and he mentioned there is a way to define wedge products using determinants. As far as I understand, given a set of vectors $x,y,z,v,u... \in \Bbb{...
3
votes
3answers
126 views

Are differential forms defined on $\Bbb{R}^{n}$

I thought $p$-forms were linear maps from $\Bbb{R}^{n} \rightarrow \Bbb{R}$. But I read something yesterday that suggested I was mistaken to think this. It seemed to be saying that $p$-forms eat $p$-...
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1answer
153 views

Wedge product of maps

If $V$ and $W$ are $\mathbb{F}$ vector spaces, A $k$-multilinear alternating map $V^k \to W$ induces a unique linear map $f: \bigwedge^k V \to W$. In the special case $W = \mathbb{F}$ and $\dim V <...
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1answer
35 views

Inverse of the Wedge of a Matrix

Let $V$ be an $n$-dimensional vector space. Then in the usual way define $\wedge^2 V$ to be the vector space spanned by the elements $v_1 \wedge v_2$ where $v_1, v_2 \in V$ such that they satisfy the ...
1
vote
0answers
115 views

Invertability of a matrix

$\newcommand{\AA}{\mathbf{A}} \newcommand{\Tr}[1]{\operatorname{Tr}\left[#1\right]}$ I have a problem that I suspect there is a “relatively” simple answer to but it is currently eluding me. I am ...
3
votes
1answer
54 views

Correspondence between wedge product and its dual

Consider the map $$ \bigwedge\nolimits^{\!k}(V^*)\to \left( \bigwedge\nolimits^{\!k}V \right)^*\\ \left( \sum_{i=1}^n (f_{1}^i \wedge \dotsb \wedge f_k^i) \right) \mapsto \left( v_1 \wedge \dotsb \...
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votes
1answer
165 views

Relationship between Levi-Civita symbol and Grassmann numbers?

The multiplication rule for Grassmann numbers $\theta_i$ is $$ \theta_i\theta_j = - \theta_j \theta_i $$ so that $\theta_i\theta_i = 0$. Multiplying three Grassmann numbers yields $$ \theta_i\theta_j\...
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vote
1answer
67 views

The Hodge dual and the Moyal product related or just notation?

The Hodge star operator $\star$ is a linear map between $\bigwedge ^pV$ and $\bigwedge ^{n-p}V$ for an inner product space $V$ of dimension $n$. So we can we write; \begin{equation} \lambda\in \...
5
votes
1answer
210 views

Use of Poincare Lemma in solving $\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$ UPDATED

You are given the following statement of the Poincaré Lemma: If $\Phi_t$ is a one-parameter family of diffeomorphisms on $\mathbb R^n$ (not necessarily a subgroup) and $X_t$ the vector field defined ...
3
votes
1answer
79 views

jordan canonical form with direct product?

I met some problems when solving Jordan canonical forms. Here are two problems: Let $f: K^3\to K^3$ be a map in JCF having the matrix: $$\begin{pmatrix} -1 & 1 & 0\\ 0 &-1&1\\ 0&...
2
votes
1answer
83 views

About the definition of the exterior power of a vector space

There are basically two different definitions of the exterior algebra and exterior powers of a vector space $V$. I here want to concentrate on the one where we define the exterior algebra as a ...