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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8
votes
3answers
361 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 ...
7
votes
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 ...
2
votes
1answer
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 ...
5
votes
1answer
101 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 ...
0
votes
1answer
134 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 ...
6
votes
3answers
429 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 ...
2
votes
3answers
195 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 ...
1
vote
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 ...
0
votes
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 ...
1
vote
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 ...
0
votes
0answers
26 views

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 $T \in \Lambda^p(V^*)$. ...
3
votes
0answers
93 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 $T \in \Lambda^p(V^*)$. ...
2
votes
1answer
387 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 ...
1
vote
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!
1
vote
1answer
207 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 ...
0
votes
0answers
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 $\phi_1, \ldots, \phi_k \in V^*$, ...
3
votes
1answer
141 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| ...
1
vote
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 ...
2
votes
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
votes
1answer
564 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 ...
6
votes
1answer
112 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 $\wedge^k(-)$ ...
2
votes
1answer
192 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 ...
4
votes
1answer
421 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, ...
0
votes
1answer
210 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 $$ ...
0
votes
1answer
152 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: $$ X=X_1\partial _x+X_2\partial _y $$ $$ Y=Y_1\partial _x+Y_2\partial _y $$ ...
1
vote
1answer
413 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 ...
5
votes
2answers
408 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 ...
2
votes
1answer
204 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 ...
6
votes
1answer
113 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 ...
1
vote
2answers
224 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 ...
1
vote
1answer
57 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
votes
1answer
137 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 ...
7
votes
1answer
183 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 ...
1
vote
1answer
61 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 ...
2
votes
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
votes
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
vote
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 ...
1
vote
1answer
141 views

Prove that if $η$ is exact, then $η∧β$ is also exact.

Prove that if $η$ is exact, then $η∧β$ is also exact. Please give a clear way to solve?
2
votes
1answer
108 views

If $ \eta $ and $ \varphi $ are closed differential forms, then prove that $ \varphi \wedge \eta $ is a closed differential form.

Let’s assume that $ \eta $ and $ \varphi $ are closed differential forms. Then how can I prove that $ \varphi \wedge \eta $ is a closed differential form as well? Please explain how to solve this ...
1
vote
1answer
381 views

Associativity property of wedge products

Regarding notation, I have a bit that says: The linear function $e_i^* \in V^*$ determined by $e_i^*(e_j) = \delta_{ij}$ form the basis of $V^*$, which is called the dual basis for $\epsilon = (e_1, ...
4
votes
1answer
296 views

Are these two definitions of exterior derivative equivalent?

I saw two definition of the exterior derivative of a $k$-form $\omega$. First definition: $$(d\omega)_p(v_0,\ldots,v_k)= \sum_{i=0}^k(-1)^id(\omega(v_0,\ldots,\hat{v_i},\ldots,v_k))_p(v_i)$$ Second ...
2
votes
1answer
494 views

Tensor and Wedge Product of Vectors

I have a little doubt about tensor product acting on vectors. I was reading Spivak's Calculus on Manifolds, and Spivak introduces the tensor product of multilinear functionals. Latter he introduces ...
3
votes
3answers
271 views

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 ...
2
votes
1answer
200 views

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 $\alpha \wedge \beta = 0$?
2
votes
1answer
59 views

Is there a smallest sub-Grassmann algebra containing a given vector in Grassmann algebra?

Let $V$ be a $d$-dimensional $\mathbb C$-vector space and the Grassmann algebra $$\mathcal G (V):=\bigoplus_{n=1}^d V^{\wedge n}$$ where $\wedge$ denotes the antisymmetric tensor product. I was ...