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.

learn more… | top users | synonyms (1)

1
vote
0answers
48 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 ...
1
vote
1answer
53 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 ...
1
vote
0answers
51 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 ...
1
vote
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
55 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
154 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 $$ ...
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
94 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
109 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
101 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 ...
6
votes
3answers
736 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 ...
9
votes
2answers
341 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 ...
4
votes
2answers
225 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
253 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 ...
4
votes
1answer
114 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
187 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
50 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
107 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 ...
0
votes
0answers
133 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 ...
3
votes
3answers
125 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 ...
1
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 ...
1
vote
1answer
34 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
109 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
155 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 $$ ...
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
209 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 ...
3
votes
1answer
78 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\\ ...
2
votes
1answer
81 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
108 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
74 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 ...
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
378 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?
1
vote
1answer
102 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
384 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 ...
6
votes
0answers
45 views

Relationship between exterior power of representation and variance?

I was reading the question: Symmetric and exterior power of representation regarding how to determine the character of an exterior power of a representation from the original representation. One of ...
0
votes
1answer
101 views

From Orthogonal vectors to Useful Bivector

If we have set of orthogonal vectors (X) can we form a set of orthogonal bivectors from that set? I am trying to find if there is a way to get 'more information' from an orthogonal matrix by some ...
7
votes
1answer
172 views

Determinant bundle of a tensor product

Let $X$ be a ringed space (for example, a scheme or a manifold). If $V$ is a locally free $\mathcal{O}_X$-module of rank $n$, then $\mathrm{det}(V) := \Lambda^n V$ is a locally free ...
6
votes
3answers
177 views

Wedge product = set intersection?

In a research article [1] I found the following formulation: The wedge product may be considered as set intersection. For example, surfaces of constant $f(x,y,z)$ and surface of constant ...
1
vote
3answers
64 views

Economically computing $d\beta$

$\displaystyle \beta = z\frac{x dy \wedge dz + y dz \wedge dx + z dx \wedge dy}{(x^2+y^2+z^2)^{2}}$ Show that $d\beta=0$. So, let $r=x^2+y^2+z^2$, $\begin{align} \displaystyle d\beta &= ...
1
vote
1answer
97 views

Identity concerning Lie derivative of $k$-form $\omega$

Let $X$ and $Y$ be vector fields on $\mathbb{R}^n$. Show that for $\omega$, a $k$-form on $\mathbb{R}^n$, $(L_XL_Y-L_YL_X)\omega=L_{[X,Y]}\omega $. I try using Cartan's magic formula and get that ...
2
votes
1answer
43 views

Different forms for the exterior power of a module

First I have defined the exterior algebra of a module $M$ as the quotient $T(M)/A(M)$ where $T(M)$ is the tensor algebra of $M$ and $A(M)$ is the ideal generated by all elements of the form $m\otimes ...
1
vote
1answer
40 views

Show that $i_Yi_Xd\omega=d\omega(X,Y)$ for $\omega$ a $1$-form

If $\omega$ is a $1$-form, how does $i_Yi_Xd\omega=d\omega(X,Y)$? I get that $d\omega$ is a 2-form. So $i_X(d\omega)=d\omega(X,v_{2})$. So how do we proceed? I dont see how the step ...
1
vote
0answers
64 views

Poincaré Lemma problems and computing contractions in an economical way

Let $x=(A,B,C,D)$ be coordinates on $\mathbb{R}^4$. $\displaystyle \beta = \frac{(AdB-BdA)\wedge(dC \wedge dD)+(dA \wedge dB)\wedge(CdD-DdC)}{(A^2+B^2+C^2+D^2)^2}$ I would like to compute ...
3
votes
1answer
94 views

Wedge product descend to the cohomology

I found this statement in Raoul Bott "Differential Forms in Algebraic Topology": "Because the wedge product is an antiderivation, it descends to cohomology." Apparently this meant to be really obvious ...