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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6
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3answers
178 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 $g(x,y,...
1
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1answer
61 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\...
2
votes
1answer
222 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 ...
1
vote
2answers
53 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$ ...
1
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1answer
164 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 ...
7
votes
1answer
165 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 ...
0
votes
1answer
36 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{\...
0
votes
2answers
84 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 ...
2
votes
2answers
89 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 ...
20
votes
2answers
1k 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 ...
1
vote
0answers
60 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 + \sum_{j=1}^{...
2
votes
1answer
64 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
votes
2answers
82 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
164 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 $\{e_1,e_2\}$...
1
vote
2answers
466 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 ...
0
votes
2answers
101 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? $(a \wedge b) \wedge c = ?$
4
votes
0answers
198 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) \...
1
vote
1answer
286 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 ...
3
votes
0answers
39 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.
5
votes
1answer
103 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, ...
1
vote
2answers
116 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 $\{X,Y\}=(XY+YX)/2$.
2
votes
2answers
157 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 $$ A(f)...
8
votes
2answers
918 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
vote
1answer
37 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 ...
2
votes
2answers
178 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 $d$ ...
1
vote
1answer
32 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.
0
votes
2answers
283 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 \\ ...
0
votes
1answer
34 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
vote
0answers
74 views

The canonical perspective on the Hodge star operator [closed]

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
votes
1answer
89 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 ...
2
votes
1answer
64 views

A basis in the $k$-th exterior power of a vector field

Definition: Let $\mathbb R^n$ be the $n$-dimensional real vector space. An exterior $k$-form call any skew-symmetric tensor on $\mathbb R^n$ of rank $k$. Denote the set of exterior $k$-forms by $E^k$. ...
4
votes
1answer
295 views

Pullback of Differential-Form

Given a differential form $$x\,dy\wedge dz-y\,dx\wedge dz+z\,dx\wedge dy$$ I am supposed to prove that the it's pullback by a linear map of determinant one leaves it invariant. For example, if $\...
1
vote
1answer
221 views

If $\omega\wedge\beta$ is exact for every closed form $\beta$, then $\omega$ is exact.

Let $\omega$ be a closed $k$-form. Then: If $\omega$ is exact, for every closed form $\beta$, the form $\omega\wedge\beta$ is exact. Proof: Let $\omega=d\alpha$. Now $d(\alpha\wedge\beta) = d\...
0
votes
2answers
171 views

wedge product of $m$ vectors in $\mathbb{R}^n$

I came across the symbol $|v_1 \wedge \dots \wedge v_m|^{-1}$ in a paper - this is the norm of the wedge product of vectors $v_k \in \mathbb{R}^n$ . I thought it's meaning was self-evident until I ...
1
vote
0answers
92 views

Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?

Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$? I know that $\Lambda(T^{*}E)=\bigoplus\Lambda^k(T^{*}E)=\Lambda^0(T^{*}E)\oplus\cdots\oplus\Lambda^n(T^{*}E)...
5
votes
3answers
235 views

Geometric introduction to exterior algebra

Could anyone point me to a geometric introduction to exterior algebra (meaning, one with a good number of figures and/or verbal descriptions of geometric objects in it)? Thanks!
1
vote
0answers
200 views

Norm inequality with wedge product

Anyone could help me to prove this following inequality? $\displaystyle\frac{||(u+v)\wedge w||}{||u+v||}\le \frac{||u\wedge w||}{||u||} +\frac{||v\wedge w||}{||v||} $ where $u\wedge v$ is the wedge ...
3
votes
2answers
113 views

Tensor products, existence of a unique linear map

Question: Given a bilinear map $B: V\times W\to X $, show there exists a unique linear map $T:V \otimes W\to X $ s.t. $B= T \circ \phi$ Background: We define $V \otimes W $ by F[ $V\...
1
vote
0answers
63 views

Grassmannian as a submanifold of the exterior product

I'm looking for a proof of the fact that if $V$ is a finitely dimensional vector space, then $G_p(V) \setminus \{0\}$ is a submanifold of $\Lambda_pV$. Here $G_p(V) = \{ v_1 \wedge ... \wedge v_p \ ...
5
votes
1answer
120 views

Exterior power of a space of maps $(\mathbb{K}^T)$

We are given a set $T \neq \emptyset, \ \ p \ge 1, \ \ p_i : T \rightarrow \mathbb{K}$ Could you help me prove that if $ \phi: (\mathbb{K}^T)^p \ni (f_1, ..., f_p) \rightarrow \rho \in \mathbb{K}^{T^...
9
votes
1answer
192 views

How to compute Ext over an exterior algebra

I found this question in several places (even on mathoverflow and mathstackexchange), but I never found a satisfying answer. Let $k$ be a field and $V$ a finite dimensional $k$-vectorspace. I ...
0
votes
1answer
154 views

Exterior power and alternating forms: explicit computations

I would like to get a more concrete understanding of a general isomorphism I have read about. I apologize if this is too basic, but I was not satisfied with the references at my disposal. Let $K$ be ...
7
votes
2answers
312 views

Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start: Calculus derivatives are good for ...
2
votes
2answers
453 views

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.
5
votes
1answer
1k views

The Hodge $*$-operator and the wedge product

On every Riemannian manifold $M$, we can consider the Hodge $*$-operator, which is characterised by the following formula: $$a \wedge *b = (a,b)\nu.$$ Here $a$ and $b$ are smooth forms on $M$, $(\ ,\ )...
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vote
0answers
36 views

Exterior product of vectors in $\mathbb{R}^4$ with integer coefficients.

Let $a, b, c, d$ be vectors with integer coordinates in $\mathbb{R}^4$ such that $k a \wedge b = c \wedge d$ for some integer $k$ and $a \wedge b \neq l v$ for any $v \in \bigwedge^2 (\mathbb{R}^4)$ ...
2
votes
3answers
157 views

$0$-th exterior power, empty product of modules and their tensor product

$\def\finiteprod#1#2{#1_{1}\times#1_{2}\times\dots\times#1_{#2}}$In Lang's algebra he defines the tensor product as a universal object in the category of multilinear maps from $\finiteprod En$ where ...
1
vote
0answers
58 views

Vanishing criterion of pure wedges

Let $R$ be a commutative ring, $M$ some $R$-module, and $m,n \in M$. Is there some criterion when $m \wedge n = 0$ in $\Lambda^2(M)$? There are some sufficient criterions, for example that $m \in \...
2
votes
1answer
48 views

Exterior product of $\Bbb Z[x,y]$

Let $R$ be the polynomial ring $\Bbb Z[x,y]$ in the variables $x,y$. If $M=R$ (so we are considering the $R$-module $R$) then $\wedge^2 (M)=0$. This was an example in Dummit and Foote's Abstract ...
1
vote
1answer
98 views

Exterior power respects $G$-action

Basic setting: Let $V$ be a $k$-vector space of finite dimension and $V^*$ its dual space. Let $\bigwedge^n V$ denote the $n$-th exterior power of $V$. Now the canonical pairing $$V \times V^{*} \...