# 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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### 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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### 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 ...