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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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!
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23 views

Why is the presheaf-“p-exterior power” of a sheaf separated?

In the first volume of the EGA ( http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1960__4_ ), p38, Grothendieck says that the presheaf-p-exterior power of a module in ringed space a separated ...
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15 views

Exterior algebra evaluation

How do you evaluate the following expression $\hat{i}\wedge\hat{j}\wedge\hat{i}$? And why does $(\hat{i}\wedge\hat{j})\wedge(\hat{i}\wedge\hat{j}\wedge\hat{k})=-\hat{k}$ philosophically?
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27 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 ...
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34 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[ ...
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1answer
41 views

exterior derivative elementary question

Let $x$ and $y$ be coordinates of a plane $\mathbb{C}^2$. Then $$ dx \wedge dy\left(\dfrac{\partial}{\partial x}\right) $$ is $dy$ or $-dy$?
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29 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 \ ...
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55 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 ...
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62 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 would ...
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43 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 ...
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80 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 ...
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exterior differential systems: understanding the antiderivation

In the book "Exterior differential systems" by Bryant, Chern, Gardner, Goldschmidt and Griffiths on page 8 the concept of a an antiderivation is introduced: An endomorphism $f$ of the additive ...
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2answers
60 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.
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Necessary and sufficient criterion for identifying decomposable k-vectors

In the exterior algebra, if $\alpha \in \Lambda^k(\mathbb{R}^n)$ is decomposable then $\alpha \wedge \alpha = 0$, but the converse is not necessarily true. Is there some sort of algebraic criterion ...
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141 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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26 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 $v \in \bigwedge^2 (\mathbb{R}^4)$ ...
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74 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 ...
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36 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 ...
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1answer
77 views

Bivector as a sum of exterior products of basis vectors.

Prove that for any 2-vector $\alpha=\Lambda^2(V)$ there is a basis $\{e_1,\ldots, e_k\}$ of $V$ such that $\alpha= e_1\wedge e_2 + e_3\wedge e_4 + \ldots +e_{k-1}\wedge e_k$, where $\wedge$ denotes ...
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1answer
37 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 ...
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66 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^{*} ...
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109 views

Hodge dual exterior derivative

The introduction of the Hodge dual to the structure of the cotangent space requires the reference to a specific basis or an inner product. I was wondering however, if the composition of hodge dual and ...
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50 views

Roots of characteristic polynomial of endomorphisms of modules under wedge products

Suppose we have a module $M$ over a ring $R$. And let $E \in End(M)$ with characteristic polynomial $f$. Let the roots of $f$ be $\lambda_1, \dots, \lambda_d$. Then $E$ corresponds to one of these ...
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278 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 ...
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1answer
22 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 ...
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179 views

Hodge-Star-Operator on arbitrary oriented basis

Assume that $V$ is oriented finite dimensional vectorspace with dimension $n$, $g \in T^0_2(V)$ a given symmetric and nondegenerate tensor. Let $\mu$ be the corresponding volume element of $V$. ...
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78 views

Kernel of the Lie bracket $[,]\colon\wedge^2\mathfrak g\to\mathfrak g$

I believe the following is probably well-known, but so far I couldn't find the answer by myself: Let $\mathfrak g$ be a real (finite-dimensional) Lie algebra, and $\wedge^2\mathfrak g$ its second ...
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77 views

Universal Property of the Exterior Algebra

Let $k$ be a field and let $A$ be a commutative algebra over $k$. I want to calculate the exterior algebra $\Lambda_A^\bullet A$. We have $\Lambda_A^0 A = \Lambda_A^1 A= A$, and $\Lambda_A^k A = 0$ ...
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75 views

How to integrate this differential form on the boundary of the cube

The setup. Assume $u = u_1+iu_2: \mathbb{R}^3 \to \mathbb{C}$ and we have the differential 1-forms $$ \star\xi=-x_2 dx_3 + x_3 dx_2 $$ and $$ u \times du = \sum_{i=1}^3 (u \times \partial_i u) dx_i = ...
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Free Graded Commutative Algebra on a Graded Vector Space

Let $V$ be a graded vector space, thought of as a collection $\{ V^n \}_{n \ge 0}$ of vector spaces. Let $V_{odd} = \bigoplus_{n \text{ odd}} V^n$ and $V_{even} = \bigoplus_{n \text{ even}} V^n$. I am ...
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96 views

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 $k\times k$ ...
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1answer
74 views

Pullback expanded form.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ According to Daniel ...
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1answer
120 views

Lie Derivative for Wedge Product of Vector Fields

I am having trouble here. The context is: Let $X$, $Y$ and $S$ be vector fields ina a manifold (we can assume it's $\mathbb{C}^2$ though I'm pretty sure this should work in any manifold), and we can ...
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1answer
63 views

Exterior Product $d\Phi_1\wedge d\Phi_2$ and spherical coordinates

One short question: If $\Phi\colon\mathbb{R}^3\to\mathbb{R}^3$, defined by $$ \begin{pmatrix}r\\\vartheta\\\phi\end{pmatrix}\mapsto\begin{pmatrix}r\sin \vartheta\cos \phi\\r\sin \vartheta ...
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3answers
146 views

index free proof of dot product of two wedge products

I am learning geometric algebra, and meet an identy of (edited according to Andrey's comments below) $$ (a\wedge b)\cdot(c\wedge d) = (a \cdot d)(b\cdot c) - (a \cdot c)(b \cdot d)$$ as in wiki ...
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1answer
57 views

Show that if $f,g\in\mathcal{S}_r (E;F)$ and $f(v,v,…,v)=g(v,v,…,v), \forall v \in E $ then $f=g$.

Give $E, F$ vectorial spaces , where $\mathcal{S}_r (E;F)$ is the vectorial space the all applications r-linear symmetrics$f$, this is, the all applications $f:E \times E \times....\times E ...
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1answer
43 views

An example of the “natural” paring $V^* \times V \rightarrow \mathbb{R}$

This so called natural paring is not natural to me at all. I am wonder if someone could give me an explicit example? I understand that $V^*$ is the dual space of $V$, and to my understanding, its ...
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2answers
56 views

If $\phi_i$s are linearly dependent, $\det [\phi_i(v_j)] = 0$ - is the proof legit?

Given $v_1, \ldots, v_k \in V$ and $\phi_1, \ldots, \phi_k \in V^*$. If $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, proof $\det[\phi_i(v_j)] = 0.$ Here $k$ is the dimension of $V$, but ...
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Did I give enough justification when I extend to $p$-dimensional?

I have proved these two exercises: (1) Suppose that $T \in \Lambda^p(V^*)$ and $v_1, \ldots, v_p \in V$ are linearly dependent. Prove that $T(v_1, \ldots, v_p) = 0$ for all $T \in \Lambda^p(V^*)$. ...
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Prove that $\phi_1 \wedge \cdots \wedge \phi_k (v_1, \cdots, v_k) = \frac{1}{k!}\det[\phi_i(v_j)].$

I have proved these two exercises: (1) Suppose that $T \in \Lambda^p(V^*)$ and $v_1, \ldots, v_p \in V$ are linearly dependent. Prove that $T(v_1, \ldots, v_p) = 0$ for all $T \in \Lambda^p(V^*)$. ...
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1answer
209 views

The wedge product

I have seen the wedge-product as being defined in differential geometry in the definition of a differential form or p-form. Now in the course we have proven the basic properties of this product and ...
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1answer
52 views

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!
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1answer
121 views

Two-form wedge product

Consider a two-form $\alpha$, then $d\alpha \wedge d\alpha$ is not necessarily zero. Is this statement true? Consider $\beta = \alpha \wedge d \alpha$. Then $d\beta = d(\alpha \wedge d ...
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141 views

Alternating tensors vs $p$-vectors

Is there a reason to differentiate between alternating tensors and $p$-vectors? More precisely, is the exterior algebra always isomorphic to the subalgebra of alternating tensors? Thanks.
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52 views

Each entry in $\phi_1, \ldots, \phi_k \in V^*$

I have always been scared by exterior algebra, which means I don't really have any background. So here's a very basic question I would like to clear out: Consider $\phi_1, \ldots, \phi_k \in V^*$, ...
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1answer
62 views

Symmetric and exterior powers of a projective (flat) module are projective (flat)

Assume that $R$ is a commutative ring with unity and $P$ a projective (flat) $R$-module. Why $\mathrm{Sym}^n(P)$ and $\Lambda^n(P)$ are projective (flat) for every $n$?
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2answers
62 views

Equality involving exterior product..

suppose you have a differential form $\omega$ writting in local coordinates as $$\omega=\sum_{i=1}^ndx_i\wedge dy_i.$$ Can anyone help me showing the following equality: $$\omega^n=n!(dx_1\wedge ...
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1answer
150 views

Multiplying vectors (answered own question)

I recently realised that asking a question and answering our own question is allowed here, so here is a question I've seen commonly on many sites: "How does one multiply two vectors?" This is very ...
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4answers
97 views

Big Greeks and commutation

Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering? Clearly if $\mathbf{x}_i$ is a matrix then: $$\prod_{i=0}^{n} \mathbf{x}_i$$ depends on the order of the multiplication. But, ...
2
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1answer
122 views

In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes?

I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y ...