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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2
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1answer
141 views

decomposable elements of $\Lambda^k(V)$

I have conjecture I have problem to prove or disprove. Let's $ w \in \Lambda^k(V)$ is k-vector. $W_w=\{v\in V: v\wedge w = 0 \}$ is k-dimensional vector space if and only if $w$ is decomposable. ...
8
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3answers
324 views

Without choosing bases, how to show that the determinant is multiplicative in this sense?

I was recently considering this statement: Let $V$ be a finite-dimensional $k$-vector space, and let $\phi:V\to V$ be an endomorphism. Suppose that $W\subseteq V$ is a subspace that is stable ...
2
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0answers
85 views

Computing wedge product of two 1-forms.

Let $L$ be a lattice in $\mathbb{C}$ and let $\pi :\mathbb{C}\to X=\mathbb{C}/L$ be a quotient map. Show that the local formula $dz$ in every chart of $\mathbb{C}/L$ is a well-defined holomorphic ...
2
votes
2answers
54 views

Compute the value of the exterior $2$-form

Compute the value of the exterior $2$-form $$\omega = (x_1 + x_2)e_1^* \wedge e_2^* + (x_2 + x_3)e_2^* \wedge e_3^* + \cdots + (x_i, x_{i+1})e_i^* \wedge ...
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1answer
30 views

2-form associated with a skew map

Given a two-form $\omega\in \Lambda^2V$ for some (say finite dimensional) vector space $V$ we may associate with $\omega$ a skew map $f_{\omega}:V\rightarrow V^*$ given by $X\mapsto \iota_X\omega$, ...
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votes
1answer
48 views

Show that $af∧bg=(ab)f∧g$

Let $v$ be vector space. For $a$ and $b$ are in IR, $f$ is in $A_{k}(V)$ and $g$ is in $A_{l}(V)$ Show that $af∧bf=(ab)f∧g$ Here Will I use the definition of wedge product? Is ti right? How to use? ...
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1answer
98 views

Use the Fundamental Theorem to deduce the formula for the area of an ellipse.

Use the Fundamental Theorem (Green's Theorem) to deduce the formula for the area of an ellipse. Hint: find a 1-form whose exterior derivative is $ dxdy $.
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votes
1answer
91 views

If $ \eta $ and $ \varphi $ are closed differential forms, then prove that $ \varphi \wedge \eta $ is a closed differential form.

Let’s assume that $ \eta $ and $ \varphi $ are closed differential forms. Then how can I prove that $ \varphi \wedge \eta $ is a closed differential form as well? Please explain how to solve this ...
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vote
1answer
131 views

Prove that if $η$ is exact, then $η∧β$ is also exact.

Prove that if $η$ is exact, then $η∧β$ is also exact. Please give a clear way to solve?
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1answer
93 views

how prove $\rho\wedge d\rho=0$ and how to show if $d(f\rho)=0$ for $f$ on $\Bbb R^{n}$ then $\rho\wedge d\rho=0$

$\def\d{\mathrm{d}}\def\R{\mathbb{R}}$Let $ρ$ be a $1$-form on $\R^{n}$ then Firstly how to prove $\rho\wedge \d\rho=0$ Secondly, how to show that If $\d(f\rho)=0$ for some nowhere vanishing smooth ...
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1answer
109 views

What is the image of the map $\hom(V,V) \to \hom(\wedge^k V,\wedge^k V)$?

The title says it all. For the uninitiated: Any map $f:V \to W$ induces a map $\wedge^k V \to \wedge^k W$ by $v_1 \wedge \cdots \wedge v_k \mapsto f(v_1)\wedge \cdots \wedge f(v_k)$, so $\wedge^k(-)$ ...
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vote
1answer
249 views

Associativity property of wedge products

Regarding notation, I have a bit that says: The linear function $e_i^* \in V^*$ determined by $e_i^*(e_j) = \delta_{ij}$ form the basis of $V^*$, which is called the dual basis for $\epsilon = (e_1, ...
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votes
1answer
236 views

Are these two definitions of exterior derivative equivalent?

I saw two definition of the exterior derivative of a $k$-form $\omega$. First definition: $$(d\omega)_p(v_0,\ldots,v_k)= \sum_{i=0}^k(-1)^id(\omega(v_0,\ldots,\hat{v_i},\ldots,v_k))_p(v_i)$$ Second ...
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vote
1answer
338 views

Tensor and Wedge Product of Vectors

I have a little doubt about tensor product acting on vectors. I was reading Spivak's Calculus on Manifolds, and Spivak introduces the tensor product of multilinear functionals. Latter he introduces ...
3
votes
3answers
199 views

Exterior Algebra as quotient

Given a vector space $W$, I understand what the tensor algebra $T(W)$ is, and I understand that the exterior algebra $\bigwedge W$ is defined as $\bigwedge W := T(W)/N$ where $N$ is the two-sided ...
6
votes
3answers
336 views

Exterior algebra of a vector bundle

Associated to any vector space $V$ is its exterior algebra $\Lambda(V)$ which has the direct sum decomposition $\Lambda(V) = \bigoplus_{i=0}^n\Lambda^i(V)$ where $n = \dim V$. My first interaction ...
2
votes
1answer
184 views

Wedge product of forms

Let $\alpha = y^2 dx + dy \in \Omega^1(R^2)$, $\beta = xy dx \wedge dy \in \Omega^2(R^2)$. Is $\alpha \wedge \beta = 0$?
2
votes
1answer
58 views

Is there a smallest sub-Grassmann algebra containing a given vector in Grassmann algebra?

Let $V$ be a $d$-dimensional $\mathbb C$-vector space and the Grassmann algebra $$\mathcal G (V):=\bigoplus_{n=1}^d V^{\wedge n}$$ where $\wedge$ denotes the antisymmetric tensor product. I was ...
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votes
2answers
194 views

$\alpha\wedge\beta = 0$ for all $\beta$ implies $\alpha = 0$ without using the Hodge dual

Let $\alpha$ be a differential $k$-form on an orientable smooth $n$-dimensional manifold. If $\alpha\wedge\beta = 0$ for every differential $(n - k)$-form $\beta$, then $\alpha = 0$ because we can ...
3
votes
0answers
136 views

Multiplication in exterior algebra

Take $V = K^{n}$. Let $\omega$ be a non-zero element of $\bigoplus_{k=1}^n \bigwedge^k V$, where we have excluded the summand $\bigwedge^0 V = K$. (1) Prove that there exists an $m > 1$ for which ...
4
votes
2answers
247 views

algebraic manipulation of differential form

Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ $(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes ...
2
votes
1answer
196 views

Wedge product with a non-degenerate form

Let $\alpha$ be a non-degenerate form in $\Lambda^k(V)$ for some vector space $V$, $\dim V = n$. (Here non-degenerate means that if $x\in V$ is nonzero, then $(y_1 , ... , y_{k-1}) \mapsto \alpha(x , ...
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vote
1answer
60 views

How can I show that $\det(v_1,v_2,\ldots,v_n)=dx_1\, dx_2\cdots dx_n(v_1,v_2,\ldots,v_n)$?

I wanted to use the definition of a wedge product which says $λ_1λ_2\cdots λ_k(v_1,v_2,\ldots,v_k)=\det(λ_i(v_j))$ with $1<i,j<k$ but I'm not sure if that even can work
7
votes
1answer
166 views

Determinant of the transpose via exterior products

Let $V$ be a finite-dimensional vector space over $F$ and let $\tau:V \to V$ be a linear operator. Here's my definition of the determinant: If $t:U \to U$ is a linear operator and $\dim(U)=n$ then ...
1
vote
0answers
124 views

Explicit computations of tensor and wedge product

Let $f\colon K^3\to K^3$ be a map in Jordan canonical form having matrix $$ f=\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0& -1 \end{bmatrix} $$ What is $f\otimes f$? What ...
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votes
5answers
4k views

Exterior Derivative vs. Covariant Derivative vs. Lie Derivative

In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them ...
2
votes
1answer
264 views

Taking the exterior derivative of a 0-form

I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative. The $f_i$ are ...
4
votes
3answers
151 views

elementary question regarding differential forms

Is it possible to give a high level explanation why changing the order of differentials will give rise to a minus sign ? I.e. why do we have $$ dx\,dt = - dt\,dx $$ (I am going to take a course on ...
5
votes
1answer
91 views

General trace relation

Let $V$ be vector space $\dim V=N$, and $A\in End(V)$. Denote $$ \wedge^k A^m(\mathbf{v}_1\wedge\dots\wedge\mathbf{v}_k)=\sum_{s_1,\dots,s_k=0,1,\sum_j s_j=m} A^{s_1}\mathbf{v}_1\wedge\dots\wedge ...
4
votes
1answer
96 views

Trace of the multiplication operator

Let $V$ be vector space, $\dim V=N$. Define the multiplication operator $L_{\mathbf{b}}$ as $L_{\mathbf{b}}:\omega\to \mathbf{b}\wedge\omega$, where $\omega\in\wedge V$ ($\wedge V$ is the entire ...
3
votes
1answer
355 views

Laplace expansion

This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix ...
5
votes
0answers
66 views

Inner product of $p$-forms [duplicate]

Possible Duplicate: Extension of Riemannian Metric to Higher Forms I have no problems with understanding the inner product of 1-forms on a Riemannian manifold. We have a metric tensor, it's ...
6
votes
1answer
162 views

$\wedge^k(V)^* \cong \mathrm{Alt}^k(V)$

Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$. I ...
5
votes
0answers
61 views

What about other symmetric functions of the eigenvalues? [duplicate]

Possible Duplicate: Identities for other coefficients of the characteristic polynomial Let $A$ be a matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. Then $\det(A) = \lambda_1 \dots ...
2
votes
1answer
432 views

Interpreting the determinant of matrices of dot products

In the Euclidean space $\mathbb{R}^n$ consider two (ordered) sets of vectors $a_1 \ldots a_k$ and $b_1 \ldots b_k$ with $k \le n$. Question What is the geometrical interpretation of ...
2
votes
0answers
148 views

Calulation of pullback of form

If $M$ is $2n+1$ dimensional manifold, and $M'= M\times \mathbb R$ Let $x_1,y_1,... x_n, y_n,t', t$ be coordiante of $M'$. With $t$ for coordinate for $\mathbb R$. Let $$ \omega= \sum_{i=1}^n ...
6
votes
1answer
677 views

wedge product of differential form

If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ ...
2
votes
2answers
171 views

Second exterior power of a complex vector space

Suppose we are given a $k$-dimensional complex vector space. Consider the second exterior power of $V$, that is $\Lambda^2V$. Denote $X=\{v_1\wedge v_2\colon v_1,v_2\in V\}$. Now I have two ...
6
votes
1answer
106 views

Wedge product of Hochschild Cohomology classes in characteristic 2

Let $A$ be a smooth commutative $k$-algebra, for $k$ a commutative ring. By the Hochschild-Kostant-Rosenberg theorem, we have that $HH^*_k(A)\cong \Lambda^* \mathrm{Der}_k(A,A)$, where ...
7
votes
1answer
171 views

A scalar product in the space of oriented volumes?

Let $L\colon \mathbb{R}^n \to \mathbb{R}^N$ be an injective linear map. By the Cauchy-Binet formula, $\det(L^TL)$ equals the sum of the squares of all minors of $L$ of order $n$: this looks just like ...
2
votes
1answer
99 views

Is $\bigwedge(V)$ self-injective?

For a vector space $V$, is the grassman algebra $\bigwedge(V)$ always an injective module over itself? Is there a proof, or even just a brief explanation?
4
votes
1answer
143 views

Are projective modules over exterior algebras of vector spaces necessarily free?

Let $E(V)$ be the exterior algebra of a vector space $V$ (I've also seen this denoted $\Lambda(V)$).Is it true that any projective $E(V)$-module is necessarily free? If it's any easier, is it at least ...
1
vote
1answer
185 views

Exterior Algebra of Self-Direct Sum

Suppose $V$ and $W$ are vector spaces and $\bigwedge V$ and $\bigwedge W$ their exterior algebras. Then it is known that $\bigwedge (V \oplus W) \simeq \bigwedge V \otimes \bigwedge W$. Now my ...
3
votes
1answer
137 views

What is the difference between $\ker( L \bigwedge L \overset{[-,-]}{\rightarrow} L )$ and $\langle a \wedge b \big| [a,b]=0\rangle$?

Let $L$ be a finite dimensional Lie algebra. We view the Lie bracket as a linear map on the exterior square: $$\pi:L \bigwedge L \rightarrow L$$ Define $$\bigwedge L := \langle a \wedge b \big| ...
2
votes
1answer
231 views

Basis of the symmetric algebra $S(M)$ given $R$-module basis of $M$ using the diamond lemma?

Over the past week, I read this secret blogging seminar post concerning the diamond lemma, which got me to reading about Bergman's paper on the diamond lemma. Now suppose you have a free $A$-module ...
11
votes
2answers
264 views

Why is it that $\det(\phi-x\text{id})=\sum_{i=0}^n (-1)^ic_ix^i$?

I'm trying to understand a certain formula for the determinant in a more general setting. Say you have a free module $M$ of rank $n$ over a (commutative) ring $R$. Let ...
2
votes
0answers
62 views

Determining explicitly the action on the exterior products of a vector space

Let $V$ be a 2-dimensional complex vector space with basis $e_1,e_2$. Consider the endomorphism $f:V\to V$ given by $f(e_1) = e_2$ and $f(e_2) = -e_1$ with matrix $$ \left( \begin{matrix} 0 & -1 ...
1
vote
0answers
52 views

Alternating forms tangential to a subspace.

Let $V$ be a finite-dimensional vector space with euclidean product, and let $U$ be a subspace. Now let $P$ be the projection of $V$ onto $U$, and let $\omega$ be any alternating multilinear $k$-form. ...
21
votes
3answers
462 views

What's the connection between derivatives and boundaries?

The (second) fundamental theorem of calculus says that $$\int_a^b f'(x) dx = f(b) - f(a)$$ which can also be stated, if one knows enough about what's coming next, as: The integral of the ...
5
votes
1answer
726 views

Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?