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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4
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66 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 $(\...
1
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1answer
89 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 ...
2
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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 \...
3
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2answers
78 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 ...
4
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1answer
253 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 ...
1
vote
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 ...
0
votes
1answer
66 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\...
3
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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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0answers
49 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
52 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
40 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 $$\...
0
votes
1answer
56 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 ...
2
votes
2answers
157 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)...
0
votes
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
96 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
112 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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votes
3answers
742 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 ...
4
votes
1answer
81 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
353 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 $w$...
5
votes
2answers
234 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 $X$. ...
8
votes
3answers
265 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
118 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
68 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
61 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
192 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
52 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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votes
0answers
136 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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vote
1answer
149 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
110 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
52 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 \...
2
votes
1answer
158 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\...
1
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
82 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 ...
1
vote
2answers
110 views

Relation between volume form and cross product

Euclidean three-dimensional space (it's simpler). Defining $\eta={e^*}^1 \wedge {e^*}^2 \wedge {e^*}^3$, with $\{{e^*}^1,{e^*}^2,{e^*}^3\}$ dual of the orthonormal basis, and indicating the classic ...
2
votes
1answer
75 views

Multilinear algebra some basics.

The wedge product of $p$ vectors in vector space $V$ is called a $p$-vector and the vector space generated by all $p$-vectors is denoted $\bigwedge^p V$ with the basis $e_I:=e_{i1}\wedge\dots\wedge e_{...
2
votes
0answers
155 views

base change of exterior powers

Let $n\geq 0$ be an integer, $R\to R'$ a ring homomorphism, and $M$ an $R$-module. Then the following holds: $$\bigl(\bigwedge^n_R M\bigr)\otimes_r R' \cong \bigwedge_{R'}^n\, (M\otimes_r R').$$ I ...
6
votes
0answers
399 views

What is the difference between tensor calculus and exterior derivative type concepts?

I am trying to clarify terms in order to help me figure out what I'd like to study. I understand that $p$-forms and $p$-vectors are used with things like wedge products, exterior algebras, and a ...
7
votes
2answers
112 views

Does a $p$-form eat $p$-vectors or $p$ number of vectors?

A bilinear form is another term for a $2$-form. So does it eat $2$ distinct vectors or a single $2$-vector?
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vote
1answer
104 views

Pull back of a vector representing a 2-form in $\mathbb R^3$

Let $\omega$ be a $2$-form defined in $\mathbb R^3$. I know that it can be represented by a vector field $\xi$ in such a way: $$ \omega_x (v,w) = \xi(x)\cdot (v\times w) $$ ($x,v,w \in \mathbb R^3$ ...
6
votes
2answers
415 views

How to visualize $1$-forms and $p$-forms?

I am having trouble understanding the common way of visualizing one-forms. Example of the visualization: On Wikipedia and in several math and physics texts books, I have come across visualizations ...
0
votes
1answer
51 views

Find a function $f$ on $\mathbb{R}^3$ such that $A^*df=0$

Let $A : \mathbb{R}^2 = \{(u,v)\} \rightarrow \mathbb{R}^3=\{(x,y,z)\}$ be given by $A(u,v)=(u,v,F(u,v))$ Find a function $f$ on $\mathbb{R}^3$ such that $A^*df=0$. I have tried $A^*df=d(...