# Questions tagged [exterior-algebra]

For questions on the exterior algebra, and related concepts such as the wedge product, the tensor algebra and differential forms.

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### Determinant Formula for Wedge Product via Universal Property of Exterior Powers

I'm currently learning about differential forms in my analysis class, and I thought I'd dig a bit more into the linear algebra of exterior powers. I've seen the universal property of $\bigwedge^k(V)$: ...
1 vote
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### How to compute exterior derivative of generalized angular form in $\mathbb R^n$

On $\mathbb R^n$, with $m\in\mathbb Z$, I want to calculate the exterior derivative of the angular form \begin{align} w=\sum_{i=1}^n(-1)^{i-1}\frac{x_i}{\|x\|^m}dx_1\cdots dx_{i-1}\hspace{0.03cm}\...
1 vote
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### exterior derivative of wedge of two one-forms

Let $\alpha, \beta$ be $k$-forms, then, where $d$ is the exterior derivative, we have $$d (\alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^k (\alpha \wedge d \beta)$$ However, in many cases, e.g. ...
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### Linear dependence and nonzero p-vectors

If a, b, and c are linearly dependent, then it's easy to show that $a\wedge b\wedge c = 0$. However, if a, b, and c are linearly independent, how do you show that $a\wedge b\wedge c \ne 0$?
1 vote
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### trace of wedge product and cyclic property [closed]

Surprisingly, while their are similar but more advanced questions on this site, I don't see any answers to the basic version I am asking herein. If I am taking the trace of a wedge product of matrices,...
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### The dot product is the norm of what vector (product)?

Let $a$, $b$, $c$ be vectors in $\mathbb{R}^3$ that form a triangle. (How this even makes sense formally, I don't know, since in a vector space all vectors are "glued" to the origin. In an ...
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### Wedge product defined on not alternating tensors?

I am currently reading Calculus on Manifolds by Spivak. In there, it defined wedge product as follows To determine the dimensions of $\Lambda^k(V)$, we would like a theorem analogous to Theorem 4-1. ...
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### Direct sum of a vector space V and a field R

What does it mean, that the exterior algebra of a vector space V over R is a superset of direct sum of real numbers and real vector space V? I thought that given vector spaces must have trivial ...
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### Integral involving Hadamard product with volume element [closed]

Consider the following integral: $$\int_{\mathbb{R}^{n\times m}}f(X)(H^T(A\odot dX)V)^{\wedge},$$ where $f:\mathbb{R}^{n\times m}\to \mathbb{R},$ $H\in O(n)$, $V\in O(m),$ $A$ is an $n\times m$ matrix ...
Let us consider vector space $$V = (\Lambda^2 (\mathbb C(t))^\times)\otimes_\mathbb Z\mathbb Q.$$ Here I consider the group $A=(\mathbb C(𝑡))^\times$ of non-zero rational functions under ...
I am stuck on a step in a proof, so I will write out the statement and the proof (it is not too long). I would like someone to explain the last 2 steps of the proof, if possible. Statement: Let $U,V \... 2 votes 1 answer 144 views ### Why is$d(xdx) = 0$? If the idea behind the exterior derivative$d$is that it tells us how quickly a$k$-form changes along every possible direction, why is$d(xdx)=0$even though$xdx$varies with$x$? I understand the ... 1 vote 1 answer 56 views ### Does an algebra over a$\mathbb{F}_2$of countable dimension have a grading where each component is finite? I have a commutative unital$\mathbb{F}_2$-algebra$Q$of countably infinite dimension, and I want to see if it has a$\mathbb{Z}$-grading where$Q_i=0$for$i<0$and$Q_0=\mathbb{F}_2$and$Q_i$... 0 votes 2 answers 55 views ### Linear transformations who keeps wedge product We have a$\mathbb K-$v.s. V. We call a linear transformation$A \in \mathscr L(V)$keeps the wedge product of two vectors of$Va$and$b$($a\wedge b$) iff$(Aa) \wedge (Ab) = a\wedge b$. I want ... 0 votes 0 answers 48 views ### Confusion with covariant derivatives with vielbeins I have some confusion regarding how the covariant derivative is defined for one forms on a manifold in the context of frames/vielbeins. I am a physics student and my reference is Sec 4.3 of the ... 0 votes 1 answer 63 views ### Explicit computation of exterior power of vector bundle I am following the excellent book by Ellinsgrud and Ottem, Introduction to Schemes. When proving that the$n$-th exterior power of the cotangent bundle of$\mathbb{P}^n$is$\mathcal{O}_{\mathbb{P}^n}$... 0 votes 1 answer 66 views ###$ω = dx_1 ∧ dy_1 + dx_2 ∧ dy_2 + dx_3 ∧ dy_3$. Prove:$∧^1 ( \mathbb{R}^6 ) \ni η \to η ∧ ω ∧ ω \in ∧^5(\mathbb{R}^6)^{*}$is a linear isomorphism. In$\mathbb{R}^6$we take variables$x_1, x_2, x_3, y_1, y_2, y_3$and a bilinear form$\omega \in \wedge^2(\mathbb{R}^6)^{*}$given by: $$\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2 + dx_3 \wedge ... 0 votes 1 answer 139 views ### Do all bivectors simplify to 2-blades in seven-dimensional space? The wedge product of two vectors \vec{v}, \vec{w}\in\mathbb{R}^{n} can be defined as an anti-symmetrized tensor product. In three dimensions, there is a correspondence between the wedge product of ... 0 votes 0 answers 26 views ### Exterior Power of a Tensor Product in case one is a line bundle I was reading this thread Exterior power of a tensor product and I found that the result cited in first answer is very useful to me, but I couldn't prove it myself and what's said there is not enough ... 9 votes 0 answers 323 views ### How big can a wedge of 2-forms be? The comass of a 2-form \alpha is the maximal value of \alpha(u,v) for a pair of unit vectors u,v. The symplectic form \alpha on \mathbb R^{2n} has the property that |\alpha^{\wedge n}| = n!... 1 vote 0 answers 31 views ### Coordinate-free proof of the identity for two times cross product operator Consider an oriented three-dimensional Euclidean space V. Let [a] be the operator of cross product by the vector a: [a] b = a \times b. It is easy to check that$$[a]^2 = a \otimes \flat a - \... 0 votes 0 answers 19 views ### Caulculation of the Supertrace of an operator I am trying to understand the following statement. We have operators$A_1 ,\ldots , A_k$for$k\leq d$. Then suppose that $$\det (x_1A_1+\cdots +x_d A_d)=a_1x_1^d+\cdots +a_d x^d+\cdots a_{12\cdots d}... 0 votes 2 answers 44 views ### Explicit calculation of exterior products Let E be a vector space. Then, for any pair of vectors e_1, e_2\in E the following holds:$$ e_1\wedge e_2 = - e_2\wedge e_1$$where$\wedge\$ denotes the exterior product. Now, in principle, this ... 