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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26 views

Proving wedge product is associative

Fix a real vector space $V$ of finite dimension. Let's denote by $\Lambda^p(V)$ the vector space of $p$-forms on $V$ (i.e. alternating $p$-tensors). Then we have the product $\wedge : \Lambda^p(V) ...
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1answer
50 views

Choice of order in the Leibniz rule is arbitrary?

One of the rules which characterizes the exterior derivative is that, for $\varphi$ a real-valued function and $\omega$ a $k$-form, we have $$d(\varphi \cdot \omega) = d\varphi \wedge \omega + ...
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35 views

Wedge product of maps: functorial vs. exterior algebra

Suppose that $V$ and $W$ are finite-dimensional vector spaces over $\mathbb{F}$. If $\varphi, \psi \in \hom(V,W)$, there are at least two interpretations of the symbol $\varphi \wedge \psi$: It is ...
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29 views

Identification between wedge product and its dual

Let $\mathbb{F}$ be a field, and let $(e_i)$ be the usual elementary basis of $\mathbb{F}^n$. Let $\varphi_{ij}: \mathbb{F}^n \wedge \mathbb{F}^n \to \mathbb{F}$ be such that $v \wedge w \mapsto ...
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39 views

Is There a Basis Free Definition of the Pfaffian

$\DeclareMathOperator{\pf}{pf}$ I recently came across a delightful fact that: The determinant of a $2n\times 2n$ skew-symmetric matrix is a the square of a certain polynomial called the pfaffian. I ...
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98 views

For $T\in \mathcal L(V)$, we have $\text{adj}(T)T=(\det T)I$.

Let $V$ be an $n$-dimensional vector space over a field of characteristic $0$. For a linear operator $T\in \mathcal L(V)$, we know that $\bigwedge^n T=(\det T)I$, where $I:V\to V$ is the identity map. ...
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29 views

Some calculations with skew forms and wedge product

I have some problems with the language of multilinear forms. I have to prove that if $dim(V)\le 3$, then every $\omega\in\Lambda^q(V^\ast)$ is such that $\omega\wedge\omega=0$. I consider the case ...
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1answer
35 views

Exercise about wedge product and multilinear forms

I'm considering $\omega\in \Lambda^{2q+1}(V^\ast)$, i.e. a multilinear skew-symmetric form. I want to prove that $\omega\wedge\omega=0$. How shall I proceed? Any suggestions? Do I have to write ...
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1answer
49 views

Question about Grassmannian, most vectors in $\bigwedge^k V$ are not completely decomposable? [closed]

My question: Is $e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 V$ not completely decomposable if $e_1$, $e_2$, $e_3$, $e_4$ is a basis for $V$?
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1answer
58 views

Star operator in the simplest form

Let $E$ together with $g$ be a inner product space(over field $\mathbb R$) , $\text{dim}E=n<\infty$ and $\{e_1,\cdots,e_n\}$ is orthonormal basis of $E$ that $\{e^1,\cdots,e^n\}$ is its dual ...
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21 views

Notation for a projection of a differential form

Let $\omega = a_1 dx_1 + a_2 dx_2 + b_1 dy_1 + b_2 dy_2$. Is there any established notation to denote a mapping that "filters out" the $dy_i$-Terms? To be more precise, I invent my own one. Assume ...
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56 views

Wedge product equality of 2n-forms in 2n+2 dimension

I have a question that seems to me clear but i couldn't prove it. I have a diffeomorphism between the cotangent bundle of the n-sphere and the complex quadric as $$T^*(S^n)\to (S^n)^{\mathbb{C}},\ ...
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1answer
30 views

The exterior algebra is a superalgebra?

Can someone explain how the exterior algebra of a vector space or a module over a commutative ring is a superalgebra? The exterior algebra has an obvious $\mathbb{Z}$-grading, but I don't see ...
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20 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 ...
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2answers
39 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 ...
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1answer
26 views

Exact sequence of vector bundles

I'am working on a shorter proof of a theorem but to manage it I need to know if a lemma is true. Conjecture: Given a manifold $M$ and an short exact sequence of vector bundles $$ 0 \rightarrow E' ...
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0answers
35 views

Geometric Interpretation of Determinant of Transpose

Below are two well-known statements regarding the determinant function: When $A$ is a square matrix, $\det(A)$ is the signed volume of the parallelepiped whose edges are columns of $A$. When $A$ is ...
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11 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
42 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 ...
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36 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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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 ($i.e.$, ...
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46 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 ...
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0answers
21 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 ...
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1answer
62 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 ...
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0answers
17 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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32 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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0answers
55 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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2answers
69 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
96 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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1answer
46 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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0answers
49 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
46 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
46 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 ...
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0answers
32 views

Exterior product of square matrix

Let $A = (a_{ij})$ be a square matrix of rank n. How to compute the exterior product $\wedge^{i} A = (b_{ij})$ for $i =1,...,n$, how to express $b_{ij} = ?$. Thank you!
3
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1answer
70 views

(Hopefully) Simple question about the exterior algebra functor

I have some (hopefully super) basic questions about the exterior algebra functor $$ \wedge:R\text{-Mod}\rightarrow R\text{-Alg}. $$ As I (think I) understand it, if one considers it as a functor ...
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0answers
34 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
31 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
49 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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2answers
138 views

Differential Forms Notation is Wrong? Confirm or deny? [closed]

Being an engineering student that just happens to have a large interest in math, I have always felt that appealing to definitions instead of intuitively understanding a concept is a mistake. A while ...
0
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1answer
43 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
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1answer
63 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 ...
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3answers
629 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
3answers
50 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 ...
4
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1answer
50 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 ...
4
votes
1answer
83 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 ...
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votes
3answers
126 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
100 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 ...
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1answer
41 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 ...
0
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0answers
41 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
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1answer
84 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 ...