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.

learn more… | top users | synonyms (1)

1
vote
2answers
48 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
vote
1answer
144 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
108 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
25 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!} ...
0
votes
2answers
73 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
81 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 ...
18
votes
2answers
747 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
45 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 + ...
2
votes
1answer
55 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
79 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
97 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 ...
1
vote
2answers
300 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
89 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? $(a \wedge b) \wedge c = ?$
3
votes
0answers
103 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
192 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 ...
2
votes
0answers
29 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.
0
votes
0answers
54 views

Let $K$ a field, $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor. Prove that $f \wedge f =0$

Let $K$ a field, with $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor ($f \in {\mathcal T}_n(V):=\Lambda^{n}(V)$), i.e., $f$ is an multilinear ...
5
votes
1answer
95 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
112 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
126 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 $$ ...
4
votes
2answers
414 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
35 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
135 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
30 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
193 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
27 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 ...
2
votes
0answers
64 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
86 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
53 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 $E^k$. ...
3
votes
1answer
188 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
111 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) = ...
0
votes
2answers
111 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
85 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 ...
5
votes
3answers
185 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
141 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
100 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[ ...
1
vote
0answers
55 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 \ ...
2
votes
0answers
57 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 ...
8
votes
1answer
150 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
84 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
217 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
288 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
648 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 $M$, $(\ ,\ ...
1
vote
0answers
34 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
132 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
52 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
45 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
80 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^{*} ...
8
votes
1answer
320 views

Effect of pullback of differential forms on an ideal

Say that the exterior differential system (EDS) corresponding to a PDE system is: $$df-f_x\,dx-f_y\,dy-f_w\,dw-f_z\,dz=0,\\ a_1\,f_x+a_2\,f_y=0,\tag{sys}$$ Of course we also require the independence ...
1
vote
1answer
26 views

To show that $\Lambda^pL(V\rightarrow W)$ and $L(\Lambda^pV\rightarrow W)$ are not necessarily isomorphic

Let $V$ and $W$ be two vector spaces. Use $L(V\rightarrow W)$ to represent the vector space of linear map from $V$ to $W$. It is proved that $\Lambda^p(V^*)\cong (\Lambda^pV)^*$, where $\Lambda$ is ...