# Tag Info

29

Short answer: the exterior derivative acts on differential forms; the Lie derivative acts on any tensors and some other geometric objects (they have to be natural, e.g. a connection, see the paper of P. Petersen below); both the exterior and the Lie derivatives don't require any additional geometric structure: they rely on the differential structure of ...

27

Since I don't have the time to give a super-detailed answer, allow me to just summarize some things that others have said, adding some additional points in the process. Hopefully this will be at least somewhat helpful. Basic differences: The exterior derivative and Lie derivative are defined in terms of the structure of a smooth manifold. By contrast, ...

17

I would have left this as a comment, but apparently I can't comment yet. To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the Wikipedia article) is a notion that generalizes to R^n and indeed any vector space---in general the output is what is called a "bivector". Now, it just so happens that in R^...

14

Maybe this is the theory that you mean: A manifold $M$ of dimension $m$ defines a $m$-current $[[M]]$, which is a functional on the space of smooth $m$-form in the following sense: $$[[M]](\omega)=\int_M\omega.$$ If $M$ is a manifold with boundary $\partial M$, then by Stoke's theorem the $m$-current $[[M]]$ and the $(m-1)$-current $[[\partial M]]$ is ...

14

Let me focus on the difference between Lie derivatives and covariant derivatives. Suppose I have a manifold with a connection $\nabla$ and a point $p$ in the manifold. Let $v$ be a vector field on $M$ and take $\xi \in T_pM$. The point to stress is that $\xi$ is not a vector field (although in practice it is often a vector field evaluated at $p$). We can ...

11

While at the vector space level, the pairing might seem slightly forced, we can derive it naturally by adding structure. Given a vector space $V$, we have a graded commutative ring $\bigwedge V = \bigoplus_i \bigwedge^i V$. Given $\phi\in V^*$, it naturally extends to a (graded) derivation $d_{\phi}$ of degree $-1$ on $\bigwedge V$. Since $d_{\phi}^2=0$ ...

11

You are looking at homological currents. Like Thomas mentioned in a comment, you should consult some textbooks on geometric measure theory. (The recent introductory book by Lin and Yang seems fairly accessible.) A bit more discussion of related ideas can be found at this MathOverflow thread.

11

Let $A$ be a commutative ring. Then $\Lambda(-)$ is a functor from $A$-modules to graded-commutative $A$-algebras which is left adjoint to the functor which takes the degree $1$ part. Because it is left adjoint, it preserves colimits, in particular coproducts. It follows $\Lambda(M \oplus N) \cong \Lambda(M) \otimes \Lambda(N)$. Looking at the $n$th degree ...

10

Yes, it is possible. (And you should find an example yourself: I will not deprive you of the joy of finding it :) )

10

If $M$ is a module over $R$, any endomorphism $\phi:M\to M$ induces an endomorphism $\Lambda^r \phi:\Lambda^rM\to \Lambda^rM$. If $M$ is free with basis $e_1,...,e_n$, then $\phi$ has a matrix $A=(a_{ij})$ in this basis. The module $\Lambda^rM$ is also free, with basis $(e_H)_{H\in\mathcal H}$ where $\mathcal H$ is the set of strictly increasing sequences ...

10

A very short answer: In finite dimensions and at least in characteristic 0, the equation $$\operatorname{d} \omega(x, y) = \omega(x) - \omega(y) - \omega([x, y])$$ allows you to define $[-,-]: V \wedge V \to V$ if you know $\operatorname{d}: V^* \to V^* \wedge V^*$ and vice versa. Furthermore, you can prove that conditions $[[x, y], z] + [[y, z], x] + [[z, ... 9 Yes, if$V$is a vector space, every projective$E(V)$-(right) module is free, because$E(V)$is a local ring and (right) projective modules over a local ring are free according to a theorem of Kaplansky. Edit Since Martin asks, here is the reason why$E(V)$is local. Consider the vector subspace$\mathfrak m=\wedge ^1V\oplus \wedge ^2V\oplus...\...

9

Well if $d\omega = 0$ then for any smooth closed curve $\gamma$ in $\mathbb{R}^2$ the area enclosed is a smooth compact manifold $M$ where $\partial M = \gamma$. From stokes theorem we have for any closed curve $\gamma$ $$\int_\gamma \omega = \int_{\partial M} \omega = \int_M d\omega = 0$$ Thus if $\gamma_1$ and $\gamma_2$ are curves with same ...

9

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...

9

It is an isomorphism, where $A$ is a commutative ring, and $V$ and $W$ are $A$-modules. I think we need $V$ and $W$ to be finitely generated and projective (but I'm not sure about this; perhaps someone can opine conerning this). See Theorem 7 here in Bergman's notes for more details. His proof includes a description of the inverse map. (In his notes, $k$ ...

9

I think the article you quote is using slightly facetious language in order to seem more interesting. In particular, when it seems to frame its question as "what are some other definitions of $i$?", it would be more honest to say What are the ways to define a multiplication operation on $\mathbb R^2$ which is compatible with normal vector addition, in a ...

8

I can construct this map abstractly, but I want to convince you that it isn't completely natural. Let's work in more generality: suppose $A \otimes B \to \mathbb{k}$ is a bilinear pairing. If I want to replace $A$ with some quotient $A/A'$, what's the natural thing to do to the pairing? If $A, B$ are finite-dimensional, then giving a bilinear pairing is ...

8

It's a superscript. The standard example is the polynomial ring $K[x_1, ... x_n]$, which is graded by total degree. That is, you can take $A^k$ to be the subspace of homogeneous polynomials of degree exactly $k$. In fact this is the free graded $K$-algebra on $n$ elements of degree $1$. People who talk about graded algebras often don't bother to point out ...

8

OK, since there's no answer provided, I'll make my comment one: As you can read here, Grassmann numbers are numbers built up from Grassmann variables $\theta_1,\theta_2,\ldots,\theta_n$ with the special property that they anticommute: $$\{\theta_i,\theta_j\}=\theta_i\theta_j+\theta_j\theta_i = 0 \; .$$ In particular, $\theta_i^2=0$. You can then study ...

8

No difference at all. I've been trying to write a little proof, but the software on this page seems to have forgotten how to write maths. :-( Anyway: I assume that by "regular cross/vector product" you mean the definition with coordinates as in Wikipedia. Try to compute both sides of your equation $(u\wedge v ) \cdot w = \det (u, v, w)$ with your ...

8

This business about working over a commutative ring $R$ is a red herring. Ultimately this is a collection of $n$ polynomial identities in $n^2$ variables $x_{ij}$ over the integers; that is, it suffices to prove this identity over $\mathbb{Z}[x_{ij}]$ as an equality of integer polynomials. But two integer polynomials are equal abstractly if and only if they'...

8

That depends on the convention in your textbook or notes. But I have seen that notation used, so I won't rule it out. "As $\omega$ is a two-form hence $\omega\wedge\omega\neq 0$" is false. Consider the two form $\mathrm{d}x\wedge \mathrm{d}y$ on $\mathbb{R}^4 = \{(w,x,y,z)| w,x,y,z\in\mathbb{R}\}$. This two form wedged with itself is zero. What you meant is ...

8

This is in general not true. One easy way to see this, is the following: Assume $V$, $W$ are $n$ and $m$ dimensional vector fields over the complex numbers, then it is fairly easy to show that $V\otimes W$ is isomorphic to $\mathbb{C}^{n\times m}$ with the following isomorphism $\phi: V\otimes W \rightarrow \mathbb{C}^{m\times n}$, which is defined as $\... 7 This is pointwise. Choose a basis of the 1-forms$\omega_1, \ldots, \omega_n.$Let$I$denote any subset of$\{1,2,\ldots,n \}$containing$k$elements. then let $$\omega_I = \omega_{i_1} \wedge \cdots \wedge \omega_{i_k}.$$ Meanwhile, let$I'$denote the subset consisting of the other$n-k$indices, that is$$I \cap I' = \{ \}, \; \; I \cup I' = \{1,... 7 1) The equation$i^2=iq+p$is not a recursive definition. It is defined as a solution of a quadratic equation. It is just like our normal definition$i^2=-1$. In this definition you can solve$i$to get the two solutions, just like when you "solve"$i^2=-1$, you get our ordinary definition of$i$. 2) This is to say, when you solve$i^2=iq+p$, i.e.,$i^2-qi-...

6

Hint: Let $\{e_1,\ldots, e_n\}$ be a basis of $V$. Then the space $\wedge^p V$ has a basis consisting of vectors of the form $e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_p}$ for some strictly increasing sequence $i_1<i_2<\ldots<i_p$ of indices. The linear mapping $\wedge^pM$ maps the vector $e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_p}$ to $M(e_{... 6 There are two views on graded rings, the view taken by algebraists and the view taken by topologists. To an algebraist, an element of the graded ring is just a formal linear combination of elements of each degree (we say an element of a single degree is homogenous). To a topolgist, you never actually add elements of different degrees. You can multiply two ... 6 I don't know how to describe the whole kernel, but I do know how to describe the generators of the kernel. Recall that$e_1\wedge e_2\dots\wedge e_k=0$if and only if$\{e_1,\dots,e_k\}$is a linearly dependent set of vectors. Furthermore, if$\{e_1,\dots e_k\}$and$\{e'_1,\dots,e'_k\}$are bases for the same subspace, then$e_1\wedge e_2\dots\...

6

Yes, this definition is chosen for a reason, as the unique solution to a pedagogical problem. Do Carmo's definition is awkward and redundant in 3 dimensions, but it is the only one among the usual definitions for the cross product that when generalized to $n$ dimensions (there is a cross-product of $n-1$ vectors in $R^n$) is rigorous, visibly basis ...

6

The best introduction I know of to the exterior product is Sergei Winitzki's free book Linear Algebra via Exterior Products. Chapter $2$ in particular I think addresses all of your questions (it is unclear how much of Chapter $1$ you need to read in order to read Chapter $2$, I guess that depends on how much linear algebra you've had).

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