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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84
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7answers
9k views

Exterior Derivative vs. Covariant Derivative vs. Lie Derivative

In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them ...
22
votes
3answers
542 views

What's the connection between derivatives and boundaries?

The (second) fundamental theorem of calculus says that $$\int_a^b f'(x) dx = f(b) - f(a)$$ which can also be stated, if one knows enough about what's coming next, as: The integral of the ...
19
votes
2answers
1k 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 ...
17
votes
4answers
8k views

Wedge product and cross product - any difference?

I'm taking a course in differential geometry, and have here been introduced to the wedge product of to vectors defined (in Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo) ...
14
votes
1answer
1k views

Wedge Product, A Novel Interpretation or Just Plain Wrong?

I have read (I think) all of the previous threads on this website (and many others) on this topic & unfortunately have not found an answer to my question. Due to the fact that I am only beginning ...
13
votes
3answers
800 views

Signs in the natural map $\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$

Let $V$ be a finite-dimensional vector space over a field $\Bbbk$. Let $V^*$ denote its dual. I strongly suspect that there is a natural map $$\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$$ that ...
12
votes
2answers
307 views

Why is it that $\det(\phi-x\text{id})=\sum_{i=0}^n (-1)^ic_ix^i$?

I'm trying to understand a certain formula for the determinant in a more general setting. Say you have a free module $M$ of rank $n$ over a (commutative) ring $R$. Let ...
12
votes
2answers
908 views

Grassmann numbers as eigenvalues of nilpotent operators?

The following question is motivated by the construction of the fermionic path/field integral, as done for example in Altland & Simons "Condensed Matter Field Theory". Consider the vector space ...
10
votes
1answer
174 views

Difference Between Tensoring and Wedging.

Let $V$ be a vector space and $\omega\in \otimes^k V$. There are $2$ ways (at least) of thinking about $\omega\otimes \omega$. 1) We may think of $\otimes^k V$ as a vector space $W$, and ...
9
votes
1answer
186 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 ...
9
votes
0answers
402 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 $V$. ...
8
votes
3answers
232 views

Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the ...
8
votes
2answers
789 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
8
votes
3answers
4k 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 ...
8
votes
3answers
399 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 ...
8
votes
2answers
204 views

For $T\in \mathcal L(V)$, we have $\text{adj}(T)T=(\det T)I$.

Let $V$ be an $n$-dimensional vector space over a field of characteristic $0$. For a linear operator $T\in \mathcal L(V)$, we know that $\bigwedge^n T=(\det T)I$, where $I:V\to V$ is the identity map. ...
8
votes
1answer
343 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 ...
8
votes
2answers
322 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 ...
7
votes
3answers
501 views

Understanding of graded algebra

I am recently learning from Loring W. Tu's An Introduction to Manifolds the concept graded algebra, which is used for introducing exterior algebra. I don't understand the following definition: An ...
7
votes
2answers
112 views

Does a $p$-form eat $p$-vectors or $p$ number of vectors?

A bilinear form is another term for a $2$-form. So does it eat $2$ distinct vectors or a single $2$-vector?
7
votes
1answer
215 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 ...
7
votes
1answer
174 views

Morphism of Exterior Algebras

Let $k$ be a field, let $V$ and $W$ be $k$-vector spaces of dimensions $n$ and $m$ respectively, and let $f:V\to W$ be a $k$-linear transformation. Let $\Lambda(V)$ and $\Lambda(W)$ denote the ...
7
votes
2answers
293 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 ...
7
votes
2answers
196 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 $n$?
7
votes
1answer
158 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 ...
7
votes
1answer
42 views

If and only if criterion for something to be a differential ideal

Let $I \subset \Omega^*(M)$ be a ($2$-sided) ideal (i.e. $I$ is a vector subspace, and for any $\alpha \in I$ and $\omega \in \Omega^*(M)$ we have $\omega \wedge \alpha \in I$). We say $I$ is a ...
7
votes
1answer
64 views

Cayley-Hamilton Theorem - Trace of Exterior Power Form

Let $V$ be an $n$-dimensional vector space over a field $F$ (the characteristic of which, for the purpose of this post, may be taken as $0$). Let $T$ be a linear operator on $V$ and $\lambda\in F$. ...
7
votes
1answer
164 views

Determinant bundle of a tensor product

Let $X$ be a ringed space (for example, a scheme or a manifold). If $V$ is a locally free $\mathcal{O}_X$-module of rank $n$, then $\mathrm{det}(V) := \Lambda^n V$ is a locally free ...
7
votes
1answer
208 views

A scalar product in the space of oriented volumes?

Let $L\colon \mathbb{R}^n \to \mathbb{R}^N$ be an injective linear map. By the Cauchy-Binet formula, $\det(L^TL)$ equals the sum of the squares of all minors of $L$ of order $n$: this looks just like ...
7
votes
0answers
290 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 ...
6
votes
3answers
725 views

Having trouble understanding generalized complex numbers

I'm reading a paper on the generalized complex numbers, but I have trouble in some of its fundamental properties. I have searched wiki but it left me none the wiser. Please see the image below, in ...
6
votes
2answers
308 views

Decomposition of product of exterior products

Suppose $V$ is a $n$-dimensional vector space. What is the kernel of $$\bigwedge^p V \otimes \bigwedge^q V\longrightarrow \bigwedge^{p+q} V$$ here $p+q \le n$.
6
votes
3answers
175 views

Wedge product = set intersection?

In a research article [1] I found the following formulation: The wedge product may be considered as set intersection. For example, surfaces of constant $f(x,y,z)$ and surface of constant ...
6
votes
1answer
583 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 ...
6
votes
2answers
234 views

$\alpha\wedge\beta = 0$ for all $\beta$ implies $\alpha = 0$ without using the Hodge dual

Let $\alpha$ be a differential $k$-form on an orientable smooth $n$-dimensional manifold. If $\alpha\wedge\beta = 0$ for every differential $(n - k)$-form $\beta$, then $\alpha = 0$ because we can ...
6
votes
1answer
1k views

wedge product of differential form

If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ ...
6
votes
3answers
593 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 ...
6
votes
1answer
178 views

$\wedge^k(V)^* \cong \mathrm{Alt}^k(V)$

Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$. I ...
6
votes
1answer
649 views

Understanding of exterior algebra

Consider the following definition from Loring W. Tu's An Introduction to Manifolds: For a finite-dimensional vector space $V$, say of dimension $n$, define ...
6
votes
1answer
123 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 ...
6
votes
2answers
360 views

How to visualize $1$-forms and $p$-forms?

I am having trouble understanding the common way of visualizing one-forms. Example of the visualization: On Wikipedia and in several math and physics texts books, I have come across visualizations ...
6
votes
1answer
1k views

Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?
6
votes
1answer
81 views

Slick proof of cross product identities

The cross product between vectors in $\mathbb{R}^3$ obeys two pleasant identities (sometimes named after Lagrange), namely $a\times(b\times c)=b(a\cdot c)-c(a\cdot b)$ $(a\times b)\cdot(c\times ...
6
votes
1answer
118 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 $\wedge^k(-)$ ...
6
votes
2answers
366 views

algebraic manipulation of differential form

Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ $(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes ...
6
votes
0answers
117 views

Why is the symbol for the exterior product a meet rather than a join?

I've moved this over to HSM. It seems odd that something that looks so much like a join [see below] would get given "the wrong symbol". It's even worse when you dualise it and get something called ...
6
votes
0answers
365 views

What is the difference between tensor calculus and exterior derivative type concepts?

I am trying to clarify terms in order to help me figure out what I'd like to study. I understand that $p$-forms and $p$-vectors are used with things like wedge products, exterior algebras, and a ...
6
votes
0answers
44 views

Relationship between exterior power of representation and variance?

I was reading the question: Symmetric and exterior power of representation regarding how to determine the character of an exterior power of a representation from the original representation. One of ...
5
votes
1answer
299 views

Can you find a 2-form not written as the wedge of two 1-forms?

I was under the impression that all 2-forms are the wedge $(\wedge)$ of two 1-forms. Is it possible to have a 2-form that you can't write as $A\wedge B$ with $A,B$ 1-forms?
5
votes
1answer
383 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 ...