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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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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 ...
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3answers
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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 $w$ ...
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2answers
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Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
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2answers
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Symmetric and wedge product in algebra and differential geometry

Which is the correct identity? $dx \, dy = dx \otimes dy + dy \otimes dx$ $~~~$or$~~~$ $dx \, dy = \dfrac{dx \otimes dy + dy \otimes dx}{2}~$? $dx \wedge dy=dx \otimes dy - dy \otimes dx$ $~~~$or$~~~$...
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2answers
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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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5answers
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Wedge product and cross product - any difference?

I'm taking a course in differential geometry, and have here been introduced to the wedge product of two vectors defined (in Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo) ...
15
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3answers
1k 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 ...
10
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1answer
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The Hodge $*$-operator and the wedge product

On every Riemannian manifold $M$, we can consider the Hodge $*$-operator, which is characterised by the following formula: $$a \wedge *b = (a,b)\nu.$$ Here $a$ and $b$ are smooth forms on $M$, $(\ ,\ )...
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2answers
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What is the relationship between the Hodge dual of p-vectors and the dual space of an ordinary vector space?

I understand what the Hodge dual is, but I can't quite wrap my head around the dual space of vector space. They seem very similar, almost the same, but perhaps they are unrelated. For instance, in $\...
10
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1answer
316 views

Geometric Intuition about the relation between Clifford Algebra and Exterior Algebra

It is common to see a relation being established between the Clifford Algebra and the Exterior Algebra of a vector space. Recently reading some texts written by Physicists I've seem applications of ...
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2answers
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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 ...
6
votes
1answer
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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)=\...
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1answer
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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 ...
11
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1answer
942 views

Deciding whether a form in the exterior power $\bigwedge^k V$ is decomposable

Let $V$ be a vector space and $\bigwedge^kV$ be the $k$th exterior power. I'm trying to find a condition that characterizes when an element $\omega \in \bigwedge^kV$ is decomposable in the sense that $...
6
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1answer
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Expressing invertible maps $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ as $\bigwedge^{d-1}A$ for some $A$

Let $V$ be a real $d$-dimensional vector space, let $\bigwedge^{d-1} V$ be its exterior power. Consider the following claim: Proposition: If $d$ is even, then every invertible linear map $\bigwedge^...
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1answer
253 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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1answer
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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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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 a ...
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1answer
108 views

Exterior Power: Find a orthornormal basis in Hilbert Space

Let $\wedge^{k} H$ with two inner products $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ where $(H, \langle \cdot, \cdot \rangle)$ is a real, n-dimensional Hilbert space ...
2
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1answer
172 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-...
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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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4answers
162 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
122 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 ...
0
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0answers
32 views

Does every subspace of the exterior algebra of dimension $>1$ contain a decomposable element?

Let $V$ be a real $n$-dimensional vector space, and let $W \le \bigwedge^k V$ be a subspace . Suppose that $\dim W \ge 2$. Does $W$ contain a non-zero decomposable element? If $\dim W=1$, then ...
12
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1answer
540 views

Difference Between Tensoring and Wedging.

Let $V$ be a vector space and $\omega\in \otimes^k V$. There are $2$ ways (at least) of thinking about $\omega\otimes \omega$. 1) We may think of $\otimes^k V$ as a vector space $W$, and $\omega\...
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3answers
683 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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2answers
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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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2answers
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How to visualize $1$-forms and $p$-forms?

I am having trouble understanding the common way of visualizing one-forms. Example of the visualization: On Wikipedia and in several math and physics texts books, I have come across visualizations ...
9
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2answers
920 views

Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start: Calculus derivatives are good for ...
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4answers
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Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$...
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2answers
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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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1answer
289 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 ...
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1answer
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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
2answers
1k 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, $\...
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2answers
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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 $V$ ...
2
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1answer
90 views

Prove that $p-$form $\omega$ on $M\times N$ is $\delta \pi(\alpha)$ iff $i(X)\omega=0$ and $L_X\omega=0$

Let $M$ be connected and let $\pi:M\times N \rightarrow N$ be the natural projection. Prove that $p-$form $\omega$ on $M\times N$ is $\delta \pi(\alpha)$ for some $p-$form $\alpha$ on $N$ if and only ...
9
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1answer
299 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 $\lambda\in F$. ...
8
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1answer
161 views

Which metrics on exterior power are induced from metrics on the base?

$\newcommand{\id}{\text{id}}$ $\newcommand{\Hom}{\text{Hom}}$ Let $V$ be a $d$-dimensional real vector space, and let $2 \le k \le d-1$. Every inner product on $V$ induces an inner product on $\...
8
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1answer
381 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
296 views

On Chevalley's linear identification of the Clifford algebra $C(\mathbf p)$ with the exterior algebra $\wedge \mathbf p$

In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following: "...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto x1$$...
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0answers
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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 \...
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votes
0answers
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Understanding the derivation of this form of the coderivative

Let $M$ be a smooth Riemannian manifold; Let $E$ be a vector bundle over $M$, equipped with a metric and a compatible connection $\nabla$. Denote by $d:\Omega^k(M,E) \to \Omega^{k+1}(M,E)$ the ...
4
votes
5answers
833 views

How to perform wedge product

I have heard all kinds of great things about Clifford/Geometric algebra, but I can't find any good resources. I have been looking EVERYWHERE for just one actual example of a wedge product being ...
3
votes
1answer
164 views

Understanding integration of k-forms

I try to wrap my head around the idea of integrating a k-form over a manifold. Loosely speaking, my intuition is, that for each point of the manifold, we do a projection onto the tangent space and ...
3
votes
1answer
240 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 $\...
2
votes
2answers
689 views

How to prove this Gram determinant

Let $E$ be an Euclidian oriented vector space of dimension $3$ and $x,y,u,w \in E$. How do we prove (without coodinates) $$ \det \begin{pmatrix} \langle x,u \rangle & \langle x,w \rangle \\ ...
2
votes
1answer
167 views

Inner Products on Exterior Powers

Let $H$ is a real, $n$-dimensional vector space. Define $\varphi \colon \operatorname{GL}(H) \rightarrow \operatorname{GL}(\wedge^{k}H)$ by $A \mapsto \wedge^{k}A$ and $\psi_{\langle \cdot, \cdot \...
6
votes
2answers
536 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 ...
5
votes
1answer
278 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
votes
2answers
137 views

Why $\bigwedge^{d-1}A=\bigwedge^{d-1}B \Rightarrow A= \pm B$

Let $V,W$ be $d$-dimensional vector spaces, and let $A,B \in \text{Hom}(V,W)$. Consider the induced maps on the exterior algebras: $\bigwedge^{d-1}A,\bigwedge^{d-1}B :\Lambda_{d-1}(V) \to \Lambda_{d-...