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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22 views

From Orthogonal vectors to Useful Bivector

If we have set of orthogonal vectors (X) can we form a set of orthogonal bivectors from that set? I am trying to find if there is a way to get 'more information' from an orthogonal matrix by some ...
7
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1answer
76 views

Determinant bundle of a tensor product

Let $X$ be a ringed space (for example, a scheme or a manifold). If $V$ is a locally free $\mathcal{O}_X$-module of rank $n$, then $\mathrm{det}(V) := \Lambda^n V$ is a locally free ...
2
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1answer
31 views

Different forms for the exterior power of a module

First I have defined the exterior algebra of a module $M$ as the quotient $T(M)/A(M)$ where $T(M)$ is the tensor algebra of $M$ and $A(M)$ is the ideal generated by all elements of the form $m\otimes ...
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1answer
26 views

Show that $i_Yi_Xd\omega=d\omega(X,Y)$ for $\omega$ a $1$-form

If $\omega$ is a $1$-form, how does $i_Yi_Xd\omega=d\omega(X,Y)$? I get that $d\omega$ is a 2-form. So $i_X(d\omega)=d\omega(X,v_{2})$. So how do we proceed? I dont see how the step ...
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3answers
53 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 &= ...
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0answers
40 views

How to phrase this identity in differential form language?

If the vector field $\mathbf B$ on $\mathbb{R}^3$ is constant, then the vector field $$ \mathbf A = \frac 1 2 \mathbf B \times \mathbf r $$ satisfies $$ \nabla \times \mathbf A = \mathbf B. $$ This ...
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0answers
36 views

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 ...
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1answer
27 views

Wedge product descend to the cohomology

I found this statement in Raoul Bott "Differential Forms in Algebraic Topology": "Because the wedge product is an antiderivation, it descends to cohomology." Apparently this meant to be really obvious ...
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87 views
+50

Use of Poincare Lemma in solving $\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$

Let $U = \mathbb{R}^3 \setminus \{(0,0,z) \}$ (ie $\mathbb{R}^3$ with the $z$-axis removed ) and consider $\beta$ on $U$ given by $\displaystyle \beta = \frac{x dy \wedge dz + y dz ...
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1answer
60 views

Identity concerning Lie derivative of $k$-form $\omega$

Let $X$ and $Y$ be vector fields on $\mathbb{R}^n$. Show that for $\omega$, a $k$-form on $\mathbb{R}^n$, $(L_XL_Y-L_YL_X)\omega=L_{[X,Y]}\omega $. I try using Cartan's magic formula and get that ...
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30 views

Bivector into orthogonal components

Suppose I have a metric $g$ and a bivector $ F $ on a four-dimensional vector space. It seems I can always decompose $ F $ into four mutually orthogonal vectors $a,b,c,d$ $$ F = a\wedge b + c\wedge d ...
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1answer
59 views

Decompose $\omega:= e_0\wedge(e_1\wedge e_2 + e_3\wedge e_4)$

$(e_1\wedge e_2 + e_3\wedge e_4)$ is well-known to be of rank 2 (can't be decomposed). On the other hand, $\omega \wedge \omega = e_1\wedge e_1 \wedge ... = 0$. According to the wikipedia article ...
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0answers
31 views

Inner product exterior algebra

I have to prove that if V is a real vectorial space (dimV=n) with inner product (.,.) then if we define $$ (v_{1}\wedge v_{2}\wedge...\wedge v_{k},w_{1}\wedge w_{2}\wedge...\wedge w_{k}) ...
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1answer
34 views

A combinatorial coefficient linked to exterior product

I am looking at the following sum $$ \sum c_1\wedge \cdots\wedge c_n $$ where the summation ranges over $c_1,\ldots,c_n$ such that each $c_i\in\{a,b\}$ and $a$ appears exactly $j$ times. Thus, using ...
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0answers
34 views

Wedge product of differential forms

I'm trying to grasp the notation and concept of wedge products(, and tensors as well). In my lecture notes, the following expansion/notation for a $(n,r)$-tensor is used: In a basis $\left\{ ...
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6 views

Condition for local Lipschitzness of pullback map for exterior forms

Given $w\in\Lambda^k(\mathbb{R}^n)$, determine the condition under which the map $T\rightarrow T^*(w)$ is locally Lipschitz, where $T\in GL_n(\mathbb{R})$ and $T^*(w)$ denotes the Pullback of $w$ by ...
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0answers
32 views

Pull back of a vector representing a 2-form in $\mathbb R^3$

Let $\omega$ be a $2$-form defined in $\mathbb R^3$. I know that it can be represented by a vector field $\xi$ in such a way: $$ \omega_x (v,w) = \xi(x)\cdot (v\times w) $$ ($x,v,w \in \mathbb R^3$ ...
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3answers
110 views

Wedge product = set intersection?

In a research article [1] I found the following formulation: The wedge product may be considered as set intersection. For example, surfaces of constant $f(x,y,z)$ and surface of constant ...
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0answers
25 views

Wedge product and derivatives

If I was given $$B=f(\phi,\psi,\theta)\,d\theta\wedge d\phi$$ for some function $f$, along with $$H_{\mu\nu\lambda}\equiv\nabla_\mu B_{\nu\lambda}+\nabla_\nu B_{\lambda\mu}+\nabla_\lambda ...
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1answer
35 views

“Adjugate” of an endomorphism of a finite-rank free module

If $M$ is a free module of finite rank $n$ over a commutative unitary ring and $a$ is an endomorphism of $M$, consider the endomorphism $\hat a$ of $M$ defined by the identity $$ x_1\wedge ...
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1answer
51 views

Identities for differential forms and vectorfields (reference request)

Recently I found the slides of a talk of J. E. Marsden, (Differential Forms and Stokes' Theorem). These slides introduce the required objects and summarize the basics of the corresponding theory. In ...
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2answers
43 views

n-form associated with a vector field with general metric

With the euclidean metric I use the musical isomorphisms to obtain $1$-form associated with a vector field, so for a vector field $\vec{F}=(f_1,f_2,f_3)$ we have $ \vec{F}^{\flat}=f_1dx+f_2dy+f_3dz$ ...
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1answer
122 views

Solving non-square linear systems with the exterior product and Cramer's rule

I'm reading the book Linear algebra via exterior products by Sergei Winitzki (which is the worst book, ever) and he shows that you can solve linear systems with a general solution with Cramer's rule ...
6
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1answer
66 views

Non-vanishing differential forms

Let $M$ be a differentiable manifold of dimension $n$. If the tangent bundle is trivial, then the cotangent bundle is trivial, and so are its exterior powers. In other words, on a parallelizable ...
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1answer
21 views

Equivalence relation of differential forms

My notes claim that $\displaystyle d\omega (x) = \frac{1}{k!} d\omega_{i_1\cdots i_k} \wedge f^{(i_1)}\wedge\cdots\wedge f^{(i_k)}$ is equivalent to $\displaystyle d\omega(x) = \frac{1}{k!} ...
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2answers
52 views

Differential identity and wedge products

Apparently $dx^{i_1} \wedge ... \wedge dx^{i_k}=d(x^{i_1}dx^{i_2}\wedge ... \wedge dx^{i_k})$ which I cannot see proved anywhere in my notes. It just stated as if it is obvious which I don't believe ...
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2answers
73 views

Part of proof that $d^2\omega=0$

The following comes from the proof in differentiable manifolds that $d^2\omega=0$. Let $f$ belong to the set of $0$-forms. From definition I have that $\displaystyle df = \frac{\partial f}{\partial ...
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1answer
408 views

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 ...
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34 views

Reference Request for differential ideals of Pfaffian forms on jet bundles

My setting is the following: Given two families of differential forms $\omega^i$ (with $i=1,...,m$) and $\mathrm d F^j$ (with $j=1...n-m$), define $$\eta^i := \sum_{j=1}^m a^i_j\omega^j + ...
2
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1answer
43 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
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2answers
72 views

Is there a way to determine the matrix of $\Lambda^k(T)$ given the matrix of $T$?

Let $T$ be an endomorphism of a finite dimensional vector space $V$. Suppose that $(v_1,\ldots v_n)$ is an ordered basis of $V$. And let $[T]$ be the matrix of $T$ with respect to this basis. Is ...
2
votes
0answers
49 views

Vector Laplace Beltrami operator on surface tangent and surface normal vector field

Consider a closed, compact, embedded surface $f:M \rightarrow \mathbb{R}^3$ and a vectorfield $X$ on the surface that can be decomposed in the surface frame basis $\{e_1,e_2,e_3\}$, where ...
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2answers
92 views

Area of projection of cube in $\mathbb{Z}^3$ onto a hyperplane

A cube with vertices $(\pm 1, \pm 1, \pm 1)$ gets projected into the plane perpendicular to vector $\mathbf{n}\in S^2$. The projection is a hexagon, how do I find the area? I think I can just ...
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2answers
77 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? $(a \wedge b) \wedge c = ?$
2
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41 views

Lie algebra: symmetric and exterior power of representation

If $\mathfrak{g}$ is a Lie algebra, $V$ and $W$ are representation of $\mathfrak{g}$ we define the action of $\mathfrak{g}$ on $V \otimes W$ in the following way: $X \cdot (v \otimes w)=(X \cdot v) ...
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1answer
93 views

What is the wedge product of multilinear forms?

The construction of $V^* \otimes V^*$ involves creating formal symbols and then adding in relations such as bilinearity by quotienting out. A bilinear form $V\times V\to F$ can be thought of as a ...
2
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0answers
24 views

For vector spaces $V,W$, probe that $\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)$ [duplicate]

For vector spaces $V,W$, probe that $\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)$. I have tried to use the universal property, but I can not create the necessary linear transformation.
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0answers
42 views

Let $K$ a field, $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor. Prove that $f \wedge f =0$

Let $K$ a field, with $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor ($f \in {\mathcal T}_n(V):=\Lambda^{n}(V)$), i.e., $f$ is an multilinear ...
4
votes
1answer
85 views

When is the rank the biggest number for which $\Lambda^m(M) \neq 0$?

I was doing some theory of Dedekind domains, and I found very useful to use the language of exterior algebra to prove the main results for finitely generated modules over Dedekind domains. I was, ...
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2answers
101 views

Idempotent operators over the exterior algebra

I am wondering if there exists a (reasonably) well-known set of operators $A_i$ over the exterior algebra such that $\{A_i,A_j\} = \frac{1}{2}(A_i +A_j)$, where $\{X,Y\}=(XY+YX)/2$.
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1answer
60 views

“Canonical” symmetrization/skew-symmetrization/alternation of multilinear functions

Is there some precise sense in which the "alternation" functor $A$ that maps a multilinear function $f\colon M^d\to N$ to the alternating multilinear function $A(f)\colon M^d\to N$ defined by $$ ...
3
votes
2answers
180 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
1
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1answer
32 views

Is the Grassmann Functor $\Lambda$ full?

Is the functor $\Lambda: \mathsf{FinDimVect}_\mathbb{R} \to \mathsf{Alg}_\mathbb{R}$ that sends a fin. dim. $\mathbb{R}$-vector space to its exterior algebra full? If not, is there a way of ...
0
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1answer
77 views

Why is the exterior derivative called exterior derivative

I am studying exterior calculus, and I think I have some grasp of what is the exterior derivative. However its name still eludes me - why is it called a derivative? Is it just because the operator $d$ ...
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1answer
26 views

Left ideals in exterior algebra $\Lambda E$ that aren't right

Let $E$ be a vector space. I'm interested in examples of left ideals in exterior algebra $ \Lambda E$ that aren't right ideals.
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2answers
123 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 \\ ...
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1answer
24 views

Writing vectors as linear combinations of bases $e_i\otimes e_j$ and $e_1\wedge e_2,e_1\wedge e_3, e_2\wedge e_3$

Write $$\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} \otimes \begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix} + \begin{pmatrix}1 \\ -1 \\ 5\end{pmatrix} \otimes \begin{pmatrix}4 \\ 0 \\ 3\end{pmatrix}$$ as linear ...
1
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0answers
45 views

The canonical perspective on the Hodge star operator

I am looking for the canonical perspective on the Hodge star operator. I want to see it done properly, not using basis for its definition, saying clearly what we assume in its definition. ...
0
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1answer
78 views

Operations in the exterior algebra. Multiplication in the direct sum of rings.

Let the exterior algebra $\Lambda(V)$ of a vector space $V$ over a field $K$ be the direct sum of the exterior powers $\Lambda^k(V),\quad k\in\overline{0,n}$. Then an element $x\in\Lambda(V)$ has the ...
0
votes
0answers
26 views

The exterior product (given by an alternative definition) is associative

Definition: Let $w^k$ and $w^l$ be a $k$-exterior form and an $l$-form on $\mathbb R^n$. For any $k+l$ vectors $x_1,\dots,x_{k+l}$ define $$w^k\wedge w^l(x_1,\dots,x_{k+l})=\sum_{\begin{matrix} ...