Questions tagged [exterior-algebra]

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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64 views

Canonical isomorphism between $\Lambda^2 V$ and $\mathbb{R}$

Let us take a two-dimensional real vector space $V$, and define $$\det \colon V \times V \rightarrow \mathbb{R} \colon (u,v) \mapsto \det(u\vert v).$$ Since this map is bilinear, the universal ...
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98 views

How do I see if $f \in V \wedge W$?

Let $V,W \leq X'$, where $X$ is a vector space and $X'$ its dual. Let $f \in X'$. How do I check if $f \in V \wedge W$? To make it concrete, Let $X$ be a real vector space with complex structure $J$, ...
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35 views

Is the image of the map $A \to \bigwedge^{k}A $ from matrices of a given rank closed?

$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ Let $V$ be a real $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed, and let $r>k$. Define $H_r=\{ ...
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49 views

Dimension of $\Lambda ^k (V)$

In Calculus on Manifolds by Spivak theorem 4-5 is as follows: The set of all $$ \varphi _{i_1} \wedge \dots \wedge \varphi _{i_k} \quad 1\leq i_1<i_2<\dots<i_k\leq n$$ is the basis for $\...
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1answer
70 views

Can we always span a decomposable form via constant coefficients?

Let $M$ be a smooth manifold of dimension $d$. Let $1 < k <d$ be an integer. Let $\omega^i$ be a local frame of the exterior power bundle $\Lambda_{k}(T^*M)$. Does there exists numbers $a_i ...
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1answer
72 views

Is the image of the map $A \to \bigwedge^{k}A $ from matrices above rank $k$ closed?

$\newcommand{\Cof}{\operatorname{cof}}$ $\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\End}{\operatorname{End}}$ $\newcommand{\GL}{\operatorname{GL}}$ Let $V$ be a real $d$-dimensional vector ...
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1answer
56 views

expansion of exterior derivative of interior product

Let $\omega$ be an 1-form and $X$ be a vectorfield. As usual $i_X \omega$ denotes the interior product and $\mathrm d$. the exterior derivative. Is there an expansion of the term $$ \mathrm d (i_X \...
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1answer
169 views

Level set representation of 1-forms in univariate functions

This question is in connection to an excellent youtube playlist by @eigenchris on Tensor Calculus. The idea of a differential $\mathrm d$ operator as turning a scalar field $f$ ($0$-form) into a ...
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1answer
40 views

${\Lambda}_k(V)$ is spanned by a basis.

Let $V$ be a vector space over $\mathbb{R}$ and let ${\Lambda}_k(V)$ be the set of all the multilinear-alternating functions from $\prod_{i = 1}^k V$ to $\mathbb{R}$. We are going to suppose that $k \...
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88 views

Is the rounded head of a tack in geometrical vectors intended?

It seems as though the thumbtack in geometrical vectors is equivalent to the wedge product of two vectors $A \times B=C:$ $$A^\mu e_\mu \wedge B^\nu e_\nu=C^{\mu\nu}e_\mu\wedge_\nu$$ which would ...
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1answer
59 views

${\Lambda}_k(V)$ is spanned by a basis $\mathcal{B}$

Let $V$ be a vector space over $\mathbb{R}$ and let ${\Lambda}_k(V)$ be the set of all the multilinear-alternating functions from $\prod_{i = 1}^k V$ to $\mathbb{R}$. We are going to suppose that $k \...
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1answer
32 views

Basic question about coefficients in external algebra

The exterior algebra formed from a vector space $V$ with vector basis $\{e_1,e_2,e_3\}$ will have basis $$\begin{align} &\Delta^0 V=\langle 1 \rangle\\ &\Delta^1 V=\langle e_1,e_2,e_3 \rangle\...
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39 views

Every $2$-Forms on $\mathbb{R}^3$ can be written as edge of $1$-Forms [closed]

Let $\omega \in \text{Alt}^{2}(\mathbb{R}^3)$ a $2$-Forms on $\mathbb{R}^3$. Show that it exists $\alpha, \beta \in \text{Alt}^{1}(\mathbb{R}^3)$ such that $\omega = \alpha \wedge \beta$.
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30 views

Calculation of line-integral

Let $$c: \left[\frac{1}{2},1\right]\times[0,1]\to\mathbb R^2 \\ c(r,t)=\big(r\cos(2\pi t),r\sin(2\pi t)\big)$$$$\omega=\frac{1}{\sqrt{x^2+y^2}}dx\wedge dy$$ Calculate $\int_c\omega$. Now we have $c^*...
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61 views

Antisymmetric Bilinear Forms and Wedge Products

I want to show that every antisymmetric bilinear form on $\mathbb{R^3}$ is a wedge product of two vectors. In other words, suppose we have a basis $\vec{u}$, $\vec{v}$, and $\vec{w}$ for $\mathbb{R}^{...
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44 views

Exterior 2-form in n-dimensional vector space

If $\omega$ is an exterior $2$-form in an $n$-dimensional vector space, prove that there exists a basis $v_1,...,v_n$ such that $\omega= v_1 \wedge v_2+...+v_{2r-1}\wedge v_{2r}$, where $r$ is the ...
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76 views

Corollary of Poincaré-Lemma

In the script of my professor, there is the following corollary of the Poincaré Lemma: If $\mathcal F$ is a vector field with rot$(\mathcal F)=0$ and $U\subset\mathbb R^n$ is an open ball, then ...
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14 views

A map from a manifold to a projective space correspondent to a subalgebra

Let $M$ be a smooth manifold of dimension $n$ and let $\mathfrak g$ be a finite dimensional subspace of $\Gamma(M,TM)$. Define $V_{\mathfrak g}:= \wedge^n\mathfrak g $. Let $N+1:=\dim V_{\mathfrak g}$ ...
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1answer
304 views

why is the exterior derivative natural?

In the study of manifolds up to a certain point, all the definitions feel natural in the sense that there's no other way to define them. I'm thinking of concepts like tangent vectors, one-forms, ...
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Is the image of the map $A \to \bigwedge^{k}A $ an embedded submanifold of $\text{GL}(\bigwedge^{k}V)$?

$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ Let $V$ be a real oriented $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed. Consider the following ...
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1answer
89 views

Is the map $A \to \bigwedge^{k}A $ an immersion?

$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ Let $V$ be a real oriented $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed. Consider the following ...
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0answers
65 views

Exterior power of the direct sum of projective modules

Can someone please point out to a (quite) detailled proof of the following result ? thanks in advance. Let $P$, $Q$ be finitely generated projective $R-$modules, of respective rank $n$ and $m$. then ...
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39 views

Calculating power of wedge products

I am working on a problem and stuck at some point. We have $\displaystyle \alpha = dx_1+ \sum_{i \geq 2}^n \xi_i \ dx_i$. I would like to calculate $(d \alpha)^{n-1}$. In other words, what is an ...
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1answer
111 views

Wedge product of three vectors

I have a basic wedge product question, I'm very new to this and have been looking at examples but still have trouble computing the following. If $u=(a_1,a_2,a_3,a_4)$ and $v=(b_1,b_2,b_3,b_4)$ and $w=(...
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1answer
40 views

Defining the exterior algebra

To define a quotient algebra $\mathcal{A} / I$, with $\mathcal{A}$ an algebra over a field $K$, I thought that $I \subset A$ is a condition which had to be required. However, I read that the exterior ...
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1answer
59 views

Find the fundamental group of (Z × Z) ∗ Z and (Z ∗ Z) × Z

this is a topology question, it's like this: Find the fundamental group of 1) (Z × Z) ∗ Z. 2) (Z ∗ Z) × Z. 3)Z ∗ · · · ∗ Z where there are n copies Here, Z stands for the integer ...
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133 views

Exterior derivative for 1-forms — derivation of the formula

I would like a help with understanding the formula for exterior derivative. Suppose that we have a differential 1-form on $\mathbb{R}^2$ given by $\omega(x)(v) = (a(x)dx + b(x)dy)(v)$. In other words $...
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Proving that $M(\omega^k)\subset V^*$ has dimension k with $M(\omega^k)=\{\eta^1\in\bigwedge^1(V^*):\eta^1\wedge\omega^k=0\}.$

For a $k$-form $0\ne \omega^k\in \bigwedge^k(V^*)$ it is $M(\omega^k)\subset V^*$ with $$M(\omega^k)=\{\eta^1\in\bigwedge^1(V^*):\eta^1\wedge\omega^k=0\}.$$ Prove that: dim $M(\omega^k)\le k$. I ...
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1answer
45 views

How to show sum of two elements is decomposable.

Let $V$ be a vector space over the field $F.$ Let $\{v_1,\ldots,v_4\}$ be linearly dependent set. Then how do I show that $w=v_1 \wedge v_2 + v_3 \wedge v_4$ is decomposable ? Note that to show $w$ ...
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1answer
195 views

Hodge Star operator is independent of the choice of basis on a finite dimensional vector space.

Let $V$ be a $m$ dimensional vector space over a field $\mathbb{F}$ and $1\leq l<m.$ Then consider the wedge product $\bigwedge^lV$ and $\bigwedge^{m-l}V.$ Fix a basis $\{e_1,\ldots,e_m\}$ of $V$ ...
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371 views

The Plucker relations are sufficient

Consider the Grassmannian of codimension-$d$ subspaces of a given vector space $E$ (over an arbitrary field), which I will define as $$ \operatorname{Gr}^d(E) = \{\text{linear surjections } \sigma: E \...
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142 views

determinant as wedge product of 1-forms

I am wondering if the determinant $\det \colon \underbrace{V \times \dots \times V}_{n-times} \to \mathbb R$ can be written as wedge product of 1-forms. (Given that $V$ is an $n$-dimensional vector ...
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1answer
238 views

wedge product/cross product

I am currently practising the wedge product, but I don't quite understand the structer overall. There is a task in my textbook marked "easy". Could anyone help me with this? I think an example would ...
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Is this a valid way to think of decomposable $k$-forms?

Let $V$ be a finite dimensional vector space and $\eta \in \Lambda^k(V^\ast)$ be decomposable and non-zero. Then there exists covectors $\omega^1,\dots,\omega^k$ such that $$\eta(v_1, \dots, v_k) = \...
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21 views

Specific example of $\mathrm{d}(F^* \alpha) = F^* (\mathrm{d} \alpha)$

I have to prove the equality for this specific example and I am lost somewhere close to the end. Let $F: \mathbb{R}^2 \to \mathbb{R}^3 : F(x^1, x^2) = (x^1- x^2, x^1 + x^2, (x^1)^2)$ and $\alpha$ be ...
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1answer
61 views

With p-forms, is $2\alpha \wedge \alpha = 0$?

I am having troubles computing the following problem. For, the 1-form and 2-form $$\eta = x^1 \mathrm{d}x^2-x^2\mathrm{d}x^1+x^3\mathrm{d}x^4-x^4\mathrm{d}x^3$$ $$\alpha = \mathrm{d}x^1 \wedge\...
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1answer
125 views

Musical isomorphisms and 1-forms

In this lecture by James Cook musical morphisms are introduced: The exterior differential of a function (function corresponding to a 0-form) results in a 1-form (otherwise referred to as a work form)....
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1answer
72 views

wedge product of differential forms.

When we speak of differential forms on a manifold $M$, we have two different visions. One is the point of view of sections of $\Lambda^rT^*M$, the other is of maps from $\mathfrak{X}(M)\times\cdots\...
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1answer
95 views

What does exterior algebra actually mean?

This question may be too basic and even silly, but I am new to exterior algebra and reading Wikipedia. Given $e_1, e_2,\cdots, e_n$ is a standard basis for a vector space $V$, what does $e_1\wedge ...
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1answer
104 views

Formula for wedge product

Suppose that $\phi_1,...,\phi_k\in V^*$ and $v_1,...,v_k\in V$, where $k=\dim V$. I am trying to show that $\phi_1\wedge\cdots\wedge \phi_k(v_1,...,v_k)=\frac{1}{k!}\det[\phi_i(v_j)]$, where $[\phi_i(...
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2answers
214 views

Why are Alternating Forms Zero when Evaluated for a Set of Linear Dependent Vectors?

I am reading a textbook on differential geometry and gauge theories and it it says off-hand that the alternating property of alternating multilinear forms implies that a form of this kind evaluated on ...
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1answer
77 views

Is the following summary of the wedge product correct?

In trying to understand the exterior or wedge product following Prof. Shifrin's series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, I would ...
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1answer
52 views

What is the value of sign($\tau$)?

Let $\tau\in S(p+q)$ be the permutation with $\tau(1)=p+1,\;\tau(2)=p+2,\;\ldots\;,\;\tau(q)=p+q$ $\tau(q+1)=1,\;\tau(q+2)=2,\;\ldots\;,\;\tau(q+p)=p$ How can I calculate the sign of this ...
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1answer
32 views

Does multilinearity hold in multiple component in tensor product and wedge product?

Let $V$ be a vector space over the field $K.$ Now consider $V \otimes V \otimes V \otimes V$ and for $x_i \in V$ let $x_1 \otimes x_2\otimes (x_3\otimes x_4 + x_5 \otimes x_6)$ be an element in the 4-...
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1answer
50 views

What do the $k$ and $k+1$ degrees in exterior derivatives count?

Intuitively, differential $k$-forms in exterior derivatives could possibly (?) be approached as $(0,k)$ tensors with a constraint (antisymmetry). A $1$-form, $\mathrm df$ will take a smooth function (...
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23 views

Adams operations and an artificial grading on K-theory

In this article by Snaith (p. 575) appears the following comment: ... these transgressive elements [...] can be located by means of the Adams operations [...]. These operate (unstably) in both the ...
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1answer
53 views

Can any form be wedged with another closed form to obtain a non zero top-form?

Let $M$ be a smooth oriented $d$-dimensional manifold, and let $\omega \in \Omega^k(M)$, $k<d$. Does there always exist a closed $(d-k)-$degree form $\eta \in \Omega^{d-k}(M)$ such that $\omega_p ...
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2answers
60 views

Sobolev approximation lifts to $L^p$ convergence of the exterior powers

I am reading the book "Geometric Function Theory and Non-linear Analysis", where the following claim is used: Let $\Omega \subseteq \mathbb{R}^n$ be a bounded open set. Let $f \in W^{1,s}(\Omega,\...
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1answer
31 views

The difference between exterior maps is bounded by the norm of the difference?

Let $V,W$ be $d$-dimensional real inner product spaces, and let $A,B:V \to W$ be linear maps. Let $\bigwedge^k A,\bigwedge^k B:\Lambda_k(V) \to \Lambda_k(W)$ be the induced maps on exterior powers. ...
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1answer
28 views

What is my mistake in calculating $**(x\wedge y)$?

According to Wikipedia, we have that the Hodge star is defined so that $$\alpha\wedge *\beta=\langle A,B\rangle\omega,$$ so it seems that in 4-dimensional space with basis $x,y,z,w$ and $\omega=x\...