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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97 views

Big Greeks and commutation

Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering? Clearly if $\mathbf{x}_i$ is a matrix then: $$\prod_{i=0}^{n} \mathbf{x}_i$$ depends on the order of the multiplication. But, ...
2
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1answer
126 views

In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes?

I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y ...
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1answer
128 views

Wedge product of vector fields

Can somebody explain me step by step how can I compute the wedge product $X\wedge Y$ of two vector fields, $X,Y$, in $\mathbb{C}^2$? We can consider $$ X=X_1\partial_x+X_2\partial_y $$ and $$ ...
2
votes
2answers
67 views

Why is $\theta \not \in C^{\infty}(S^1)$?

Why is $\theta \not \in C^{\infty}(S^1)$? I know that since $\int_{S^1} d\theta = 2\pi$ then $d\theta$ is not exact. Thus since $d(\theta)=d\theta$, $\theta$ must not be $C^{\infty}$ but it seems ...
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1answer
108 views

Complicated calculation on Lie bracket and wedge product

I'm in the following situation: let's assume I am given the following vector fields in $\mathbb{C}^2$, not identically zero: $$ X=X_1\partial _x+X_2\partial _y $$ $$ Y=Y_1\partial _x+Y_2\partial _y $$ ...
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vote
1answer
56 views

Change of base rings for exterior algebra

This may be not a good question. But I really get tough. I am studying basic knowledge about homological algebras and I am dealing with Koszul's Complex and Hilbert's Syzygy Theorem. At the very ...
1
vote
1answer
187 views

Anti-derivation of an exterior algebra

Let $K$ be a commutative ring. $M$ is a free $K$-module of rank $m$. $E(M)$ is the exterior algebra defined by $M$. $\iota$ is a $K$-homomorphism from $M$ to $E(M)$ such that $\iota(x)=-x$. Since ...
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0answers
26 views

I get the value of my $2$ form to be $0$.

Let $\omega = (x_1 + \cdots +x_n) \sum_{i<j} dx_i \wedge dx_j$ be a $2$-form on $\mathbb{R}^n, \omega \in \Omega (\mathbb{R}^n)$. Compute the function $$\omega \left( ...
0
votes
1answer
77 views

wedge product and change of variables

The question is: Let $\phi:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a $C^1$ map and let $y=\phi(x)$ be the change of variables. Show that d$y_1\wedge...\wedge ...
4
votes
2answers
220 views

If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable.

I am trying to prove the following from a book I am reading through. Thm: If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable. Note ...
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1answer
115 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 ...
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1answer
141 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 ...
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2answers
142 views

Computing wedge products

Compute $\omega = (e_1^* + e_2^* + \cdots+ e_n^*) \wedge (e_1^* + e_2^*) \wedge (e_1^* + e_3^*) \wedge \cdots \wedge(e_1^* + e_n^*)$ in the standard form. I first thuoght I'd pick a value from ...
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vote
1answer
51 views

Putting the wedge product in standard/normal form

I have to compute the wedge product of $$(e_1^* + ze_2^*) \wedge (e_2^* + ze_3^*) \wedge \cdots \wedge (e_{n-1}^* + ze_n^*) \wedge (e_n^* + ze_1^*),$$ and then put it in normal/standard form. So I ...
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1answer
98 views

Elements of $\wedge^2V$ expressible in the form $v_1\wedge v_2$

If $V$ is a complex vector space, then an element $w\in \wedge^2V$ is of the form $v_1\wedge v_2$ for some $v_1,v_2\in V$ iff $w\wedge w=0$ in $\wedge^4V$. Could anybody give some intuition/show why ...
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1answer
50 views

Basis for the space of $p$-forms

According to the book I read, a general $p$-form can be written as: $$\omega=\omega_{a_1\ldots a_p} dx^{a_1}\wedge\ldots\wedge dx^{a_p},\hspace{0.5cm} a_1>a_2>\ldots>a_p$$ where I have used ...
2
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1answer
119 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. ...
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votes
3answers
304 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 ...
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0answers
79 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
52 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
29 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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1answer
43 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
96 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
84 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
122 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?
0
votes
1answer
90 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
104 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
169 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, ...
4
votes
1answer
203 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
222 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
170 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
263 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
172 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
55 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 ...
5
votes
2answers
176 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
128 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
215 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
171 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
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votes
1answer
146 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 ...
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vote
0answers
110 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 ...
44
votes
5answers
3k 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
228 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
2answers
113 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
87 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
89 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
314 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
65 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
153 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
57 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 ...