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18

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 ...


15

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, ...


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 ...


13

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 ...


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.


10

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


10

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 $d_{\phi}^2=0$ ...


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 ...


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 ...


8

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 ...


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 ...


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 ...


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 ...


7

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 ...


7

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 ...


7

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 ...


6

Let me just retell a part of the story in slightly different language — maybe it will become less mysterious then. The story starts with a module $M=\mathbb C[\xi]$ (free supercommutative algebra with one odd generator — aka exterior algebra on one generator) over a Clifford algebra $A=\mathbb C[\xi,\frac\partial{\partial\xi}]$ (“odd differential ...


6

The canonical map $\psi\colon V^k \to \bigwedge^k V$ given by $\psi(v_1, \ldots, v_k) = v_1 \wedge \cdots \wedge v_k$ is alternating. If you have an element $g \in (\bigwedge^k V)^*$ then $g \circ \psi\colon V^k \to \mathbf R$ is also alternating, and I believe that this assignment $g \mapsto g \circ \psi$ gives you an inverse to $\Phi$. This is simpler ...


6

We have that $$d(\varphi \wedge \eta) = d\varphi \wedge \eta + (-1)^{\deg(\varphi)} \varphi \wedge d\eta = 0 + 0 = 0.$$ Hence $\varphi \wedge \eta$ is closed as well.


6

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' = ...


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 ...


6

If I wanted $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$ I would write it as $$\prod_{i=0}^{n} \mathbf{x}_{n-i}$$


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 ...


6

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, ...


5

This is a long comment rather than a complete answer. [Updated with a complete answer below.] $\wedge^k V$ is a simple example of a Schur functor. You might know of the classification of irreducible representations of the symmetric group in terms of partitions. Namely, given a partition $\lambda=(\lambda_1,\cdots,\lambda_k)$ where $\lambda_1\geq ...


5

I suspect that what you mean is $a\wedge b = 1/2(a\otimes b-b\otimes a)$, in which case $a\otimes b$ is the tensor product of $a$ and $b$. $a\otimes b$ is an element of a new vector space that is higher-dimensional than the vector space containing $a$ and $b$. You can think about it as just a formal construction (basically it is mostly like the pair ...


5

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 ...


5

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).


5

You don't need to choose any particular basis; all you need is the fact that $\wedge V$ is graded and $L_b$ raises the degree by $1$, so if you choose any basis that respects the gradation, then $L_b$ sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish.


5

If you want an explicit coboundary, take the boundary of the function $g$ sending $[x^a y^b]$ to $x^{a-2} y^b$ times the image of $a(a-1)/2$ in $\mathbb{F}_2$. It works because the cup square of $\partial / \partial x$ sends $[x^i y^j | x^k y^l]$ to $ikx^{i+k-2}y^{j+l}$, and $dg ( [x^i y^j | x^k y^l]) = x^i y^j g(x^k y^l) - g(x^{i+k}y^{j+l}) + g(x^i y^j) ...



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