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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6
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3answers
725 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
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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 ...
8
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2answers
322 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
215 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
228 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
67 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
186 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 ...
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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
104 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
128 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
145 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
33 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
103 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 ...
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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
149 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
207 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
73 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
365 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
97 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
357 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
50 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
44 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
163 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
175 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
39 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
63 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
92 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 ...
2
votes
0answers
63 views

Bivector into orthogonal components

Suppose I have a metric $g$ and a bivector $ F $ on a four-dimensional vector space. It seems I can always decompose $ F $ into four mutually orthogonal vectors $a,b,c,d$ $$ F = a\wedge b + c\wedge d ...
1
vote
1answer
113 views

Decompose $\omega:= e_0\wedge(e_1\wedge e_2 + e_3\wedge e_4)$

$(e_1\wedge e_2 + e_3\wedge e_4)$ is well-known to be of rank 2 (can't be decomposed). On the other hand, $\omega \wedge \omega = e_1\wedge e_1 \wedge ... = 0$. According to the wikipedia article ...
0
votes
1answer
39 views

A combinatorial coefficient linked to exterior product

I am looking at the following sum $$ \sum c_1\wedge \cdots\wedge c_n $$ where the summation ranges over $c_1,\ldots,c_n$ such that each $c_i\in\{a,b\}$ and $a$ appears exactly $j$ times. Thus, using ...
1
vote
0answers
114 views

Wedge product of differential forms

I'm trying to grasp the notation and concept of wedge products(, and tensors as well). In my lecture notes, the following expansion/notation for a $(n,r)$-tensor is used: In a basis $\left\{ ...
7
votes
1answer
158 views

Non-vanishing differential forms

Let $M$ be a differentiable manifold of dimension $n$. If the tangent bundle is trivial, then the cotangent bundle is trivial, and so are its exterior powers. In other words, on a parallelizable ...
4
votes
2answers
399 views

Lie Derivative for Wedge Product of Vector Fields

I am having trouble here. The context is: Let $X$, $Y$ and $S$ be vector fields ina a manifold (we can assume it's $\mathbb{C}^2$ though I'm pretty sure this should work in any manifold), and we can ...
8
votes
2answers
784 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
1
vote
1answer
161 views

Solving non-square linear systems with the exterior product and Cramer's rule

I'm reading the book Linear algebra via exterior products by Sergei Winitzki (which is the worst book, ever) and he shows that you can solve linear systems with a general solution with Cramer's rule ...
5
votes
1answer
298 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 $(\wedge)$ of two 1-forms. Is it possible to have a 2-form that you can't write as $A\wedge B$ with $A,B$ 1-forms?
0
votes
1answer
34 views

Writing vectors as linear combinations of bases $e_i\otimes e_j$ and $e_1\wedge e_2,e_1\wedge e_3, e_2\wedge e_3$

Write $$\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} \otimes \begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix} + \begin{pmatrix}1 \\ -1 \\ 5\end{pmatrix} \otimes \begin{pmatrix}4 \\ 0 \\ 3\end{pmatrix}$$ as linear ...