Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [exterior-algebra]

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

The exterior algebra is the quotient algebra of the tensor algebra of the associated vector space.

It uses the 'wedge' ($\wedge$) binary operator, often called the wedge product, and is alternating, anticommutative and graded.

1248 questions
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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 algebra ideal generated by $\{v \otimes w + w \otimes…
nigel
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Wedge product of 0-form with 1-form

What is the wedge product $\wedge$ of a $0$-form $f(x_1,...,x_n)$ with a $1$-form $\displaystyle\sum_{i=1}^{n} a_i dx_i$? According to the theory, it should be a (0+1=1)-form.
Andreas K.
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Can you find a 2-form not written as the wedge of two 1-forms?

I was under the impression that all 2-forms are the wedge $(\wedge)$ of two 1-forms. Is it possible to have a 2-form that you can't write as $A\wedge B$ with $A,B$ 1-forms?
JimJones
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Hodge star - Duality: Why is $\star{\star\alpha}=\pm\alpha$?

According to Wikipedia, we have $\star{\star\alpha}=\pm\alpha$ (the sign depends on the dimension of the vector space, the signature of the bilinear form and the degree of $\alpha$). I guess we could prove that by considering an orthonormal basis…
Filippo
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Intuitively, does the wedge product describe the oriented area?

Is the following interpretation correct: Given two vectors in $\mathbb{R}^n$, their wedge product describes an oriented area with magnitude equal to the area of the parallelogram formed by the two vectors and orientation perpendicular to both?
Michael Stachowsky
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Characterization of rank in an exterior algebra

The wikipedia page on exterior algebras makes the following reasonable sounding statement (I paraphrase): Let $V$ be a complex vector space and consider the second exterior power $\bigwedge^2 V$. By the rank of $\alpha \in V$, we mean the smallest…
Oliver
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Transformation rule for a wedge product

Suppose two sets of covectors on a vector space $V, \beta^1,...\beta^k $ and $\gamma^1,...,\gamma^k,$ are related by $$\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$$ where $i=1,...,k$, for a $k\times k$ matrix $[a^i_j]$. I want to show $$\beta^1…
Sepideh Bakhoda
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Definition of the exterior algebra

In Tu's An Introduction to Manifolds; an exterior algebra is defined as follows: For a finite dimensional vector space $V$, say of dimension $n$, define $$A_*(V) = \bigoplus_{k=0}^\infty A_k(V) = \bigoplus_{k=0}^n A_k(V)$$ where $A_k(V)$ is the…
QCD_IS_GOOD
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Why are $2$-covectors on $\mathbb{R}^3$ decomposable?

How do you show that every $2$-covector $\omega\in\Lambda^2((\mathbb{R}^3)^\ast)$ is decomposable i.e. that $$\omega=u\wedge v$$ for some $u,v\in(\mathbb{R}^3)^\ast$? In general we have $$\omega=\omega_{12}e^1\wedge e^2+\omega_{13}e^1\wedge…
user224776
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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 constructing an arbitrary Grassmann algebra homomorphism…
user154917
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Wedge product of forms

Let $\alpha = y^2 dx + dy \in \Omega^1(R^2)$, $\beta = xy dx \wedge dy \in \Omega^2(R^2)$. Is $\alpha \wedge \beta = 0$?
Petr Hradil
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Wedge Product Algebra

If $A′,B′,C′$ are the midpoints of $BC, CA, AB$ respectively, then show that $$4A′ ∧ B′ ∧ C′ = A ∧ B ∧ C.$$ So to begin with I have $4A' = 2BC, B' = 1/2(CA), C' = 1/2(AB)$ and therefore $4A′ ∧ B′ ∧ C′ = 2(B-C) ∧ 1/2(A-C) ∧ 1/2(B-A)$, then I try to…
Mcarr
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About the definition of the exterior power of a vector space

There are basically two different definitions of the exterior algebra and exterior powers of a vector space $V$. I here want to concentrate on the one where we define the exterior algebra as a quotient of the tensor algebra. Now for this, i have…
Mekanik
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Unusual (to me) operator used for Inner Product

I have been doing a self-study of Differential Forms and Exterior Calculus using the book "Applied Exterior Calculus" by Dominic Edelen (Dover Publication). I ran across an operator that apparently represents the inner product of a differential form…
K7PEH
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For a $k$ form $\eta$ and a $l$ form $\omega$, what is $d(\eta\wedge\omega)?$

For a $k$ form $\eta$ and a $l$ form $\omega$, what is $d(\eta\wedge\omega)?$ Thank you very much for your help and guidance!
WishingFish
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