# 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 power of multilinear functions applied to linearly dependent vectors is zero

I'm working on a homework problem, and we are to show that if $T \in \wedge^p V^*$, and $v_1,\ldots,v_p$ are linearly dependent, then $T(v_1,\ldots,v_p) = 0$. What I've got so far: I understand that ...
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### The kernel of an antiderivation on an exterior algebra

This is a simple algebraic question I feel I should be obvious, but maybe isn't. Let $d'\colon V \twoheadrightarrow W$ be a surjective linear map of finite-dimensional vector spaces over a field of ...
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### Differential Geometry-Hodge Star

The Hodge star is given by $$*(dx^{i_1}\wedge dx^{i_2}\wedge....\wedge dx_{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2....i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}$$ The question ...
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### What is the anticommutator of the interior product and codifferential (adjoint of exterior derivative)?

What is $\eta=i_X \delta + \delta i_X$ acting on differential forms? Here $\delta$ is the usual Hodge adjoint of the exterior derivative and $i_X$ is contraction of a form with the vector field $X$. ...
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### Relationship between exterior power of representation and variance?

I was reading the question: Symmetric and exterior power of representation regarding how to determine the character of an exterior power of a representation from the original representation. One of ...
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### 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 ...
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### 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 $\mathcal{O}_X$-...
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### 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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### Use of Poincare Lemma in solving $\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$ UPDATED

You are given the following statement of the Poincaré Lemma: If $\Phi_t$ is a one-parameter family of diffeomorphisms on $\mathbb R^n$ (not necessarily a subgroup) and $X_t$ the vector field defined ...
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### 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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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 $g(x,y,z)... 1answer 92 views ### “Adjugate” of an endomorphism of a finite-rank free module If$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 ax_2\... 1answer 835 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 ... 2answers 100 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$... 1answer 340 views ### Solving non-square linear systems with the exterior product and Cramer's rule I'm reading the book Linear algebra via exterior products by Sergei Winitzki (which is the worst book, ever) and he shows that you can solve linear systems with a general solution with Cramer's rule ... 1answer 456 views ### Non-vanishing differential forms Let$M$be a differentiable manifold of dimension$n$. If the tangent bundle is trivial, then the cotangent bundle is trivial, and so are its exterior powers. In other words, on a parallelizable ... 1answer 76 views ### Equivalence relation of differential forms My notes claim that$\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!} \frac{\...
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 ...
### 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 ...