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)

3
votes
2answers
101 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 = ...
6
votes
0answers
263 views

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 ...
3
votes
1answer
230 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 $k\times k$ ...
2
votes
1answer
105 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 ...
4
votes
2answers
365 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 ...
0
votes
1answer
187 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 ...
2
votes
3answers
208 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
60 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
64 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
29 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 $T \in \Lambda^p(V^*)$. ...
4
votes
0answers
115 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
454 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
242 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 ...
3
votes
1answer
168 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.
0
votes
0answers
56 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 $\phi_1, \ldots, \phi_k \in V^*$, ...
7
votes
2answers
179 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 $n$?
1
vote
2answers
81 views

Compute the $n$th exterior power of a differential $2$-form

Suppose you have a differential form $\omega$ written 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
1answer
208 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
4answers
101 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
votes
2answers
280 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 ...
0
votes
1answer
236 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 $$ ...
2
votes
2answers
76 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 ...
0
votes
1answer
171 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
99 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 ...
1
vote
1answer
499 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 ...
0
votes
1answer
111 views

Wedge product and change of variables

The question is: Let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map and let $y = \phi(x)$ be the change of variables. Show that $$dy_1\wedge\dots\wedge dy_n = ...
5
votes
2answers
489 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 ...
5
votes
1answer
539 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 ...
2
votes
1answer
242 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 ...
1
vote
2answers
254 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
60 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
149 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 ...
1
vote
1answer
70 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 ...
6
votes
2answers
300 views

Decomposable elements of $\Lambda^k(V)$

I have a conjecture. I have a problem proving or disproving it. Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if ...
8
votes
3answers
397 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 ...
2
votes
0answers
101 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
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
108 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 $.
2
votes
1answer
114 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
150 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?
0
votes
1answer
101 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 ...
6
votes
1answer
117 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
492 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, ...
5
votes
1answer
361 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
603 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
312 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 ...