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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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^{*} ...
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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 ...
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103 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 ...
3
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1answer
161 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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2answers
157 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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2answers
94 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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1answer
191 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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3answers
363 views

Without choosing bases, how to show that the determinant is multiplicative in this sense?

I was recently considering this statement: Let $V$ be a finite-dimensional $k$-vector space, and let $\phi:V\to V$ be an endomorphism. Suppose that $W\subseteq V$ is a subspace that is stable ...
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2answers
3k views

What are “Super Numbers”?

I'm reading Hyperspace by Michio Kaku and in the chapter on SuperGravity "Super Numbers" are mentioned and are described as a number system where for any super number $a$, $a*a=-a*a$. I was wondering ...
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99 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
102 views

General trace relation

Let $V$ be vector space $\dim V=N$, and $A\in End(V)$. Denote $$ \wedge^k A^m(\mathbf{v}_1\wedge\dots\wedge\mathbf{v}_k)=\sum_{s_1,\dots,s_k=0,1,\sum_j s_j=m} A^{s_1}\mathbf{v}_1\wedge\dots\wedge ...
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1answer
138 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
441 views

Exterior algebra of a vector bundle

Associated to any vector space $V$ is its exterior algebra $\Lambda(V)$ which has the direct sum decomposition $\Lambda(V) = \bigoplus_{i=0}^n\Lambda^i(V)$ where $n = \dim V$. My first interaction ...
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3answers
196 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
59 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
48 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
62 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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0answers
96 views

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
396 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
64 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
209 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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53 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
142 views

What is the difference between $\ker( L \bigwedge L \overset{[-,-]}{\rightarrow} L )$ and $\langle a \wedge b \big| [a,b]=0\rangle$?

Let $L$ be a finite dimensional Lie algebra. We view the Lie bracket as a linear map on the exterior square: $$\pi:L \bigwedge L \rightarrow L$$ Define $$\bigwedge L := \langle a \wedge b \big| ...
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2answers
75 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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2answers
73 views

Why is $\theta \not \in C^{\infty}(S^1)$?

Why is $\theta \not \in C^{\infty}(S^1)$? I know that since $\int_{S^1} d\theta = 2\pi$ then $d\theta$ is not exact. Thus since $d(\theta)=d\theta$, $\theta$ must not be $C^{\infty}$ but it seems ...
2
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1answer
571 views

“And” symbol? Wedge product in a surface integral? — Is this a typo, or did I miss an important lecture?

This is the question I got on my final assignment (Calculus III): Evaluate the surface integral \begin{equation} \int \int_S xy \; \; dy\wedge dz - yz \; \; dz\wedge dx + xz \; \; dx\wedge ...
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1answer
114 views

What is the image of the map $\hom(V,V) \to \hom(\wedge^k V,\wedge^k V)$?

The title says it all. For the uninitiated: Any map $f:V \to W$ induces a map $\wedge^k V \to \wedge^k W$ by $v_1 \wedge \cdots \wedge v_k \mapsto f(v_1)\wedge \cdots \wedge f(v_k)$, so $\wedge^k(-)$ ...
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1answer
193 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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1answer
434 views

Proving the Poincare Lemma for $1$ forms on $\mathbb{R}^2$

I am trying to prove the Poincare Lemma for $1$ forms on $\mathbb{R^2}$. So I said that if I doing this, I start of with $$\omega = f_1(x_1,x_2) dx_1 + f_2(x_1,x_2)dx_2.$$ First thing I want to ...
4
votes
4answers
100 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, ...
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1answer
211 views

Wedge product of vector fields

Can somebody explain me step by step how can I compute the wedge product $X\wedge Y$ of two vector fields, $X,Y$, in $\mathbb{C}^2$? We can consider $$ X=X_1\partial_x+X_2\partial_y $$ and $$ ...
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1answer
160 views

Complicated calculation on Lie bracket and wedge product

I'm in the following situation: let's assume I am given the following vector fields in $\mathbb{C}^2$, not identically zero: $$ X=X_1\partial _x+X_2\partial _y $$ $$ Y=Y_1\partial _x+Y_2\partial _y $$ ...
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1answer
428 views

Anti-derivation of an exterior algebra

Let $K$ be a commutative ring. $M$ is a free $K$-module of rank $m$. $E(M)$ is the exterior algebra defined by $M$. $\iota$ is a $K$-homomorphism from $M$ to $E(M)$ such that $\iota(x)=-x$. Since ...
1
vote
1answer
82 views

Change of base rings for exterior algebra

This may be not a good question. But I really get tough. I am studying basic knowledge about homological algebras and I am dealing with Koszul's Complex and Hilbert's Syzygy Theorem. At the very ...
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votes
2answers
420 views

If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable.

I am trying to prove the following from a book I am reading through. Thm: If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable. Note ...
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1answer
208 views

Computation of the hom-set of a comodule over a coalgebra: $Ext_{E(x)}(k, E(x)) = P(y)$.

First of all, since every other book somehow mentions that this is trivial, I apologize if it turns out that I am just misunderstanding something in the definitions. So here goes: The motivation for ...
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1answer
115 views

Wedge product of Hochschild Cohomology classes in characteristic 2

Let $A$ be a smooth commutative $k$-algebra, for $k$ a commutative ring. By the Hochschild-Kostant-Rosenberg theorem, we have that $HH^*_k(A)\cong \Lambda^* \mathrm{Der}_k(A,A)$, where ...
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2answers
227 views

Computing wedge products

Compute $\omega = (e_1^* + e_2^* + \cdots+ e_n^*) \wedge (e_1^* + e_2^*) \wedge (e_1^* + e_3^*) \wedge \cdots \wedge(e_1^* + e_n^*)$ in the standard form. I first thuoght I'd pick a value from ...
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1answer
58 views

Putting the wedge product in standard/normal form

I have to compute the wedge product of $$(e_1^* + ze_2^*) \wedge (e_2^* + ze_3^*) \wedge \cdots \wedge (e_{n-1}^* + ze_n^*) \wedge (e_n^* + ze_1^*),$$ and then put it in normal/standard form. So I ...
4
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1answer
139 views

Elements of $\wedge^2V$ expressible in the form $v_1\wedge v_2$

If $V$ is a complex vector space, then an element $w\in \wedge^2V$ is of the form $v_1\wedge v_2$ for some $v_1,v_2\in V$ iff $w\wedge w=0$ in $\wedge^4V$. Could anybody give some intuition/show why ...
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1answer
187 views

Determinant of the transpose via exterior products

Let $V$ be a finite-dimensional vector space over $F$ and let $\tau:V \to V$ be a linear operator. Here's my definition of the determinant: If $t:U \to U$ is a linear operator and $\dim(U)=n$ then ...
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1answer
62 views

Basis for the space of $p$-forms

According to the book I read, a general $p$-form can be written as: $$\omega=\omega_{a_1\ldots a_p} dx^{a_1}\wedge\ldots\wedge dx^{a_p},\hspace{0.5cm} a_1>a_2>\ldots>a_p$$ where I have used ...
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0answers
94 views

Computing wedge product of two 1-forms.

Let $L$ be a lattice in $\mathbb{C}$ and let $\pi :\mathbb{C}\to X=\mathbb{C}/L$ be a quotient map. Show that the local formula $dz$ in every chart of $\mathbb{C}/L$ is a well-defined holomorphic ...
2
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2answers
58 views

Compute the value of the exterior $2$-form

Compute the value of the exterior $2$-form $$\omega = (x_1 + x_2)e_1^* \wedge e_2^* + (x_2 + x_3)e_2^* \wedge e_3^* + \cdots + (x_i, x_{i+1})e_i^* \wedge ...
1
vote
1answer
104 views

Use the Fundamental Theorem to deduce the formula for the area of an ellipse.

Use the Fundamental Theorem (Green's Theorem) to deduce the formula for the area of an ellipse. Hint: find a 1-form whose exterior derivative is $ dxdy $.
1
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1answer
36 views

2-form associated with a skew map

Given a two-form $\omega\in \Lambda^2V$ for some (say finite dimensional) vector space $V$ we may associate with $\omega$ a skew map $f_{\omega}:V\rightarrow V^*$ given by $X\mapsto \iota_X\omega$, ...
0
votes
1answer
50 views

Show that $af∧bg=(ab)f∧g$

Let $v$ be vector space. For $a$ and $b$ are in IR, $f$ is in $A_{k}(V)$ and $g$ is in $A_{l}(V)$ Show that $af∧bf=(ab)f∧g$ Here Will I use the definition of wedge product? Is ti right? How to use? ...
0
votes
1answer
98 views

how prove $\rho\wedge d\rho=0$ and how to show if $d(f\rho)=0$ for $f$ on $\Bbb R^{n}$ then $\rho\wedge d\rho=0$

$\def\d{\mathrm{d}}\def\R{\mathbb{R}}$Let $ρ$ be a $1$-form on $\R^{n}$ then Firstly how to prove $\rho\wedge \d\rho=0$ Secondly, how to show that If $\d(f\rho)=0$ for some nowhere vanishing smooth ...