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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41 views

Geometric meaning of Berezin integration

Berezin integration in a Grassmann algebra is defined such that its algebraic properties are analogous to definite integration of ordinary functions: linearity (taking anticommutativity into account), ...
3
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2answers
56 views

Reference needed: Exterior covariant calculus

I would like to understand Cartan's formalism of exterior covariant calculus. I think it could be useful for some calculations in physics (But If I am wrong here and it's only good for abstract ...
3
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1answer
69 views

Different definitions for exterior derivative? [duplicate]

Let $\omega \in \Omega^k(M)$ and $X_i \in \Gamma(TM)$. In Spivak Volume I page 213 the exterior derivative is given invariantly by the formula \begin{align*}d \omega(X_0, \ldots, X_k) = & \sum_i (-...
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1answer
89 views

Interior product of a differential form with respect to a commutator of two vector fields

I've been working my way through Nakahara's book "Geometry, topology & physics" and I've reached chapter 5 section 4 in which he discusses the interior product and lie derivative of differential ...
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0answers
15 views

ℂ version of alternating property

The alternating property says that if arguments $x_i$ and $x_j$'s places are switched, then the $\mathtt{output} \mapsto - \mathtt{output}$. Is there an extension where the index set of the arguments,...
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1answer
46 views

Is matrix algebra a special case of Grassmann (exterior) algebra, and if so what is more general case?

Just a little question. I only recently heard about Grassmann algebra while reading a book on the history of vector algebra and quaternions. I still don't understand what does exterior product mean ...
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1answer
155 views

Vectors, Forms, Multivectors, and Tensors

In researching some of the ways that vectors (and vector fields) generalize I find that there are apparently many different objects that generalize them -- matrices, differential forms/ covectors, ...
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1answer
64 views

Finding a form to solve a wedge product equation

We start out with $(\mathbb{R}^{2n},\omega_0)$, the "standard" symplectic manifold. We define $\Omega=\omega_0\wedge\dotso\wedge\omega_0$, i.e. $\Omega$ is the product of $\omega_0$ with itself $n$ ...
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1answer
18 views

Exterior algebra on a smooth manifold

We have the exterior 1-form on $\mathbb{R}^3$ given by $\alpha=x dy\wedge dz-ydz\wedge dx\in\Omega^2(\mathbb{R}^3)$. I'm trying to get a explicit expression of $\alpha$ restricted to the surface $xy=1$...
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114 views

Koszul sign convention and symmetric group action on the graded n-th tensor product

Let $V_\bullet = (V_k)_{k \in \mathbb{Z}}$ and $W_\bullet = (W_k)_{k \in \mathbb{Z}}$ be two graded vector spaces on 0 caracteristic field. We define the tensor product of $V_\bullet$ by $W_\bullet$ ...
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1answer
120 views

What do we call the result of wedging together the columns of a matrix?

We can wedge together the columns of a square matrix to compute its determinant. More generally, the exterior product of the columns of a $b \times a$ matrix tells us the determinant of each $a \times ...
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1answer
27 views

Notation in exterior algebra

I am following a course on introduction to manifolds by myself and I got stuck by a notation I don't understand. It defines the permutation group $S_n$, and then the signature of a permutation as ...
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27 views

Regarding Adjugate of a Module Homomorphism

$\newcommand{\adj}{\text{adj}}$ I am trying to understand the concept of adjugate of an endomorphism of free module of finite rank as discussed here in Section 8. Let $M$ be a free module over a ring ...
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2answers
64 views

Is exterior algebra an example of an algebra over a field?

An algebra over a field is a pair $(V,\cdot)$ where $V$ is a vector space over a field $F$ with a bilinear product $\cdot:V\times V\to V$. In the Szekeres' book on Mathematical Physics, he explains ...
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2answers
55 views

Pullback of a linear map on a 2-form.

I am having a bit of trouble understanding a homework question and was seeking some clarification. Note, I have edited this question after I worked a couple of things out. Given a 2-form $v=dx_1 \...
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1answer
49 views

Exterior powers of vector space and kernel

Let $E$ be a vector space and $A$ a subspace of $E$. Let $q$ be a positive integer. Then we can define a subspace $\Lambda^q A$ of the $q$-th exterior power of $E$ by $$ \Lambda^qA=span\{\ a_1\wedge\...
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1answer
221 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
61 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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0answers
51 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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2answers
68 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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85 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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2answers
228 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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1answer
42 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
57 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
72 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
73 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 basis(...
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1answer
33 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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0answers
71 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}},\ (x,...
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1answer
55 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 where ...
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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
votes
2answers
64 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 $\theta$...
3
votes
1answer
49 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' \...
5
votes
2answers
155 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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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
53 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 $2$-...
4
votes
1answer
62 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 ...
2
votes
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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0answers
116 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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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 ...
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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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0answers
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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votes
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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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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 $(\...
3
votes
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
votes
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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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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votes
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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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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^...
3
votes
1answer
113 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 ...