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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350 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 $w$...
8
votes
2answers
912 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
86
votes
7answers
9k 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 ...
13
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3answers
820 views

Signs in the natural map $\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$

Let $V$ be a finite-dimensional vector space over a field $\Bbbk$. Let $V^*$ denote its dual. I strongly suspect that there is a natural map $$\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$$ that ...
6
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1answer
659 views

Understanding of exterior algebra

Consider the following definition from Loring W. Tu's An Introduction to Manifolds: For a finite-dimensional vector space $V$, say of dimension $n$, define $$A_*(V)=\oplus_{k=0}^{\infty}A_k(V)=\...
4
votes
1answer
62 views

Is there a nowhere $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$?

Let $M$ be a closed $3$-manifold, and let $\xi$ be a $2$-dimensional subbundle of $TM$. Is there a nowhere zero $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a ...
2
votes
1answer
113 views

Pullback expanded form.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ According to Daniel Robert-...
2
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1answer
71 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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0answers
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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 $\alpha=\...
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Symmetric and wedge product in algebra and differential geometry

I have been struggling with this issue for a while (and asked a similar question here), but still not found a satisfying answer. The question boils down to: which is the correct identity? $dx \, dy ...
8
votes
2answers
227 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. ...
3
votes
2answers
94 views

Determinant from Paul Garrett's Definition of the Characteristic Polynomial.

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\End}{End}$ On pg. 390 of Paul Garrett's notes on Algebra, a definition for the characteristic polynomial is given, which I discuss here. Let $V$ ...
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votes
1answer
68 views

Cayley-Hamilton Theorem - Trace of Exterior Power Form

Let $V$ be an $n$-dimensional vector space over a field $F$ (the characteristic of which, for the purpose of this post, may be taken as $0$). Let $T$ be a linear operator on $V$ and $\lambda\in F$. ...
4
votes
1answer
112 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 ...
3
votes
3answers
340 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 ...
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1answer
134 views

Wedge product and change of variables

The question is: Let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map and let $y = \phi(x)$ be the change of variables. Show that $$dy_1\wedge\dots\wedge dy_n = (\operatorname{det}D\...
7
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1answer
210 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 ...
4
votes
1answer
129 views

Trace of the $k$-th Exterior Power of a Linear Operator

Let $V$ be an $n$ dimensional vector space over a field $F$ and $T$ be a linear operator over $V$. Assume that the characteristic of $F$ is not $2$. Definition. Consider the map $f_1:V^n\to \...
7
votes
1answer
217 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 ...
6
votes
1answer
616 views

Proving the Poincare Lemma for $1$ forms on $\mathbb{R}^2$

I am trying to prove the Poincare Lemma for $1$ forms on $\mathbb{R^2}$. So I said that if I doing this, I start of with $$\omega = f_1(x_1,x_2) dx_1 + f_2(x_1,x_2)dx_2.$$ First thing I want to ...
5
votes
1answer
165 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 ...
5
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0answers
69 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 \...
4
votes
1answer
1k views

Is there a formula for the determinant of the wedge product of two matrices?

I was going over the Wikipedia page for exterior products of vector spaces and we can define the determinant as the coefficient of the exterior product of vectors with respect to the standard basis ...
2
votes
1answer
255 views

Computation of the hom-set of a comodule over a coalgebra: $Ext_{E(x)}(k, E(x)) = P(y)$.

First of all, since every other book somehow mentions that this is trivial, I apologize if it turns out that I am just misunderstanding something in the definitions. So here goes: The motivation for ...
2
votes
1answer
64 views

Some wedge product computation

I want to calculate $w\wedge w$ for $w=\sum a_{ij} e_i\wedge e_j$ where the sum is over all $1\leqslant i<j\leqslant 4$. Is there a neat way to do it without writing all the terms out? What about ...
3
votes
1answer
207 views

What is a more formal name for the wedgey group?

The rule $(v_1,w_1)⋅(v_2,w_2)=(v_1+v_2,w_1+w_2+(v_1∧v_2))$ defines a group structure on the vector space $V⊕(V∧V)$ whenever $V$ is itself a vector space over some field $F$. What is a more common ...
3
votes
1answer
64 views

Poincare's lemma for 1-form

Let $\omega=f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz$ be a differentiable 1-form in $\mathbb{R}^{3}$ such that $d\omega=0$. Define $\hat{f}:\mathbb{R}^{3}\to\mathbb{R}$ by $$\hat{f}(x,y,z)=\int_{0}^{1}{(f(...
3
votes
1answer
254 views

Transformation rule for a wedge product

Suppose two sets of covectors on a vector space $V, \beta^1,...\beta^k $ and $\gamma^1,...,\gamma^k,$ are related by $$\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$$ where $i=1,...,k$, for a $k\times k$ ...
2
votes
1answer
39 views

Prove the exterior derivative of the following (n-1) form is zero

Let $\omega(x)=\frac{1}{{\parallel x \parallel}^n}\displaystyle\sum_{i=1}^{n}(-1)^{i-1}x_{i} dx_{1} \wedge \dots \wedge \widehat{dx_{i}} \wedge \dots \wedge dx_{n}$ be a differential $(n-1)$ form on $\...
1
vote
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 ...
1
vote
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 $\...
1
vote
0answers
50 views

Simple properties of wedge product [closed]

How to prove a) $\omega \wedge \eta =(-1)^{kl}\eta\wedge\omega, \omega$ is $k$-tensor and $\eta$ is $l$-tensor. b)$f^*(\omega \wedge \eta)=f^*(\omega)\wedge f^*(\eta)$ where $f:V\rightarrow W$ ...
1
vote
1answer
129 views

Interior product and exterior product

I have seen around the internet that this should hold: $$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$ where $X$ is a vector field, $\alpha$ a $k$-form, $\...
1
vote
2answers
65 views

If $\phi_i$s are linearly dependent, $\det [\phi_i(v_j)] = 0$ - is the proof legit?

Given $v_1, \ldots, v_k \in V$ and $\phi_1, \ldots, \phi_k \in V^*$. If $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, proof $\det[\phi_i(v_j)] = 0.$ Here $k$ is the dimension of $V$, but I ...
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 &= d(...
0
votes
2answers
101 views

Exterior product in 3 dimension

I'm curious about the exterior product in $3$ dimensions. Is it the same as the cross product? If it is, does anybody here have an idea about its calculation? $(a \wedge b) \wedge c = ?$
0
votes
1answer
61 views

Problem proving Cartan's identity

There is a famous identity stating that, if $X$ is a field and $\omega$ a form, then: $$\mathcal{L}_X\omega=\iota_Xd\omega+d\iota_X\omega.$$ I'm trying to prove it. Thanks to Anthony Carapetis, I ...