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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5k 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 ...
21
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3answers
484 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 ...
17
votes
2answers
601 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 ...
14
votes
4answers
6k 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 Perdig√£o 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 ...
12
votes
2answers
850 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 ...
11
votes
3answers
611 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 ...
11
votes
2answers
266 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 ...
8
votes
3answers
99 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
3answers
354 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
1answer
313 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 ...
7
votes
3answers
314 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
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 ...
7
votes
2answers
175 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
1answer
181 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
2answers
153 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 ...
7
votes
1answer
105 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
185 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
1answer
122 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 ...
7
votes
0answers
318 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 $V$. ...
6
votes
2answers
292 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 $p+q \le n$.
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votes
3answers
128 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
2answers
212 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
796 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
1answer
169 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
110 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
85 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?
6
votes
3answers
398 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
90 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 ...
6
votes
2answers
122 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 $n$?
6
votes
1answer
110 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
0answers
115 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
35 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 ...
6
votes
0answers
191 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 ...
5
votes
2answers
333 views

Using the notation of wedge product to solve a linear system of equations

I am trying to solve a problem that seems like a standard idea from linear algebra but with a the notion of wedge product and exterior algebra added it gets more complicated for someone who isn't ...
5
votes
2answers
393 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
3answers
172 views

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!
5
votes
1answer
151 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 ...
5
votes
1answer
91 views

When is the rank the biggest number for which $\Lambda^m(M) \neq 0$?

I was doing some theory of Dedekind domains, and I found very useful to use the language of exterior algebra to prove the main results for finitely generated modules over Dedekind domains. I was, ...
5
votes
2answers
300 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 ...
5
votes
1answer
97 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 ...
5
votes
1answer
887 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?
5
votes
2answers
205 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 ...
5
votes
0answers
67 views

Inner product of $p$-forms [duplicate]

Possible Duplicate: Extension of Riemannian Metric to Higher Forms I have no problems with understanding the inner product of 1-forms on a Riemannian manifold. We have a metric tensor, it's ...
5
votes
0answers
64 views

What about other symmetric functions of the eigenvalues? [duplicate]

Possible Duplicate: Identities for other coefficients of the characteristic polynomial Let $A$ be a matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. Then $\det(A) = \lambda_1 \dots ...
4
votes
3answers
590 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 ...
4
votes
4answers
99 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, ...
4
votes
1answer
257 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?
4
votes
2answers
295 views

Exterior power “commutes” with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ ...
4
votes
1answer
149 views

Are projective modules over exterior algebras of vector spaces necessarily free?

Let $E(V)$ be the exterior algebra of a vector space $V$ (I've also seen this denoted $\Lambda(V)$).Is it true that any projective $E(V)$-module is necessarily free? If it's any easier, is it at least ...