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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 ...


5

This is a special case of a more general phenomenon. Let $G$ be a finite group acting on a vector space $W$ over a characteristic $0$ field. One can define two spaces. The invariants $W^G=\{w\in W\,|\, gw=w\text{ for all }g\in G\}$, and the coinvariants $W_G=W/\langle g\cdot w-w\rangle$. Here $\langle g\cdot w-w\rangle$ is the subspace spanned by vectors of ...


4

We have $y = (y_1, \dots, y_n) = (\phi_1, \dots, \phi_n)$ so \begin{align*} dy_1\wedge\dots\wedge dy_n &= d\phi_1\wedge\dots\wedge d\phi_n\\ &= \left(\frac{\partial \phi_1}{\partial x_1}dx_1 + \dots +\frac{\partial \phi_1}{\partial x_n}dx_n\right)\wedge\dots\wedge\left(\frac{\partial \phi_n}{\partial x_1}dx_1 + \dots +\frac{\partial ...


4

The lateness of this answer gives a new meaning to the word "overdue". In any case, I noticed this question while browsing this website and I thought I'd answer it because that's what you do when you're faced with a question right? I'm not sure if this is what you're looking for; it's basically equivalent to your basis-dependent proof above and it's based ...


3

The right place to start is with determinants. The determinant of a 2 by 2 matrix can be thought of geometrically as the (signed) area of the parallelogram formed by the column vectors. The exterior product is a way of formalizing this intuition. The expression $dx^0\wedge dx^1\wedge...\wedge dx^{n-1}$ is by definition the volume element. The wedge ...


3

You can prove this by proving the contrapositive. Suppose $\omega$ is not decomposable. That means that $\omega$ necessarily must be the sum of two (and only two) decomposable terms, like $\omega = x \wedge y + z \wedge w$. Consider what happens if you add a term like $y \wedge z$. You should realize that you can lump that into one of the two terms and ...


3

Suppose $\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$ then consider, \begin{align} \beta^1 \wedge \cdots \wedge \beta^k &= \biggl[ \sum_{j_1=1}^k a^1_{j_1} \biggr] \gamma ^{j_1} \wedge \cdots \wedge \biggl[\sum_{j_k=1}^k a^k_{j_k} \gamma ^{j_k} \biggr]\\ &= \sum_{j_1=1}^k \sum_{j_k=1}^k a^1_{j_1} \cdots a^k_{j_k} \gamma ^{j_1} \wedge \cdots \wedge ...


3

As you suspect the answer is yes, and you basically explain why it is so. If you are not convinced you could add the following details: your PDE $a_1 f_x + a_2 f_y=0$ is first order with the variable $f$ the dependent variable while $x,y,w,z$ are the independent variables. Standard coordinates in first jet space are $x,y,w,z,f,f_x,f_y,f_w,f_z$ while the ...


3

I think that even if it's not written explicitly anywhere, the $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ convention is the most predictable and sensible. I've never seen the distinction made explicit, since in most circumstances the operation involved is commutative. I did see somewhere on m.SE someone suggest ...


2

If your elements commute with eachother, then there is no need for an ordering in the case of finite sums/products. In the non-commutative case things are more complicated. Anyhow, IMO there is no need for an ordering if the sum/product doesn't depend on the order. And this covers many non-commutative cases too. Otherwise, it is clear that one should ...


2

By definition, if $V$ is a vector space (forget about it having a basis) over a field $k$, then the dual space $V^*$ is the space of linear functionals $f\colon V\to k$. Thus the most natural thing to do if you have a linear functional $f\in V^*$ and a vector $v\in V$, is to take $f(v)$. This gives you the natural pairing $V^*\times V\to k$, $(f,v)\mapsto ...


2

Just compute the dimensions. Suppose $\dim V = n$ and $\dim W = m$. On the one hand, $\Lambda^p L (V \to W)$ has dimension $\frac{(n m)!}{p! (n m - p)!}$; on the other hand, $L (\Lambda^p V \to W)$ has dimension $\frac{n! m}{p! (n - p)!}$. Of course, these two quantities are equal when $m = 1$, but already for $m = 2$ we have $\frac{(2 n)!}{p! (2 n - p)!}$ ...


2

Dot product (Scalar Product) The dot product, you could say, very hand-wavily measures both the overall size of 2 vectors and how parallel they are. The dot product is related to the magnitudes and angles of the two vectors by: $$\vec a\cdot\vec b=||\vec a||\mbox{ }||\vec b||\cos\theta$$ So, if the two vectors are orthogonal, their dot product is 0. If ...


2

Since both sides are multilinear (linear in each variable when the rest are fixed), it is enough to check it for combinations of a basis. There are not too many distinct cases: e.g. when $a=b$ (and say, $=e_i$), then we get $0$ on both sides. More cases to check: $\ (e_i\land e_j)\cdot (e_i\land e_j)$, $\ (e_i\land e_j)\cdot (e_i\land e_k)$, $\ (e_i\land ...


2

Sure. You can argue it pretty well just from symmetry. As was said by Berci, it has to be linear in each term, and it has to be antisymmetric under exchange of $ a $ and $ b $ or $c$ and $d$. I'm not sure if you can show that the above is the only such relation. It certainly make me feel more comfortable with the identity at least. But, regardless, you can ...


2

A motivating example with $n=2$. Let $\omega=dx_1\wedge dy_1 + dx_2\wedge dy_2$ be our 2-form. Then $\omega\wedge\omega=(dx_1\wedge dy_1 + dx_2\wedge dy_2)\wedge (dx_1\wedge dy_1 + dx_2\wedge dy_2)=dx_1\wedge dy_1\wedge dx_1\wedge dy_1+ dx_2\wedge dy_2\wedge dx_2\wedge dy_2+ dx_1\wedge dy_1\wedge dx_2\wedge dy_2+dx_2\wedge dy_2\wedge dx_1\wedge dy_1$, ...


2

Good question! Here's a start. The ordinary derivative in one-variable calculus is a Lie derivative along a special vector field on $\mathbb{R}$; in particular, it is not a special case of the exterior derivative. The exterior derivative is instead some kind of "universal derivative": it records all of the information you would need to determine the ...


2

Hints: $d\alpha$ is a 3-form. The commutativity relationship between a $k$-form $\omega$ and a $l$-form $\eta$ is $$ \omega \wedge\eta = (-1)^{kl}\eta \wedge \omega .$$ Also, $$ d(\omega \wedge\eta) = d\omega \wedge\eta + (-1)^{k}\omega \wedge d\eta. $$ The solutions should follow immediately from these properties (together with $d^2=0$). (For the ...


2

I suppose generally a $p$-form is a sum of such terms, but if we can understand how one such element pulls-back then linearity extends to $\sum_{i_1, \dots , i_p}\alpha_{i_1,\dots , i_p} dy^{i_1} \wedge \cdots \wedge dy^{i_p}$. That said, to calculate $\gamma$ you just have to sort out the sign needed to arrange the indices on the wedge of $\omega_{\alpha} ...


2

If $E$ is a complex vector bundle of rank $r$, its first Chern class is equal to the first chern class of its top exterior product: $$c_1(E)=c_1(\wedge ^r E)$$ This is extremely useful since the first (and only!) chern class of a line bundle is generally easy to compute. For example on a compact Riemann surface or on a smooth projective curve, the line ...


2

Hint: $$d\Phi_1:=\frac{\partial \Phi_1}{\partial r}dr+\frac{\partial \Phi_1}{\partial \theta}d\theta+\frac{\partial \Phi_1}{\partial \phi}d\phi= \sin\theta\cos\phi dr+r\cos\theta\cos\phi d\theta-r\sin\theta \sin\phi d\phi,$$ and similarly for the other components $\Phi_2$, $\Phi_3$ of $\Phi$, i.e. $$d\Phi_2:=\frac{\partial \Phi_2}{\partial ...


2

Your expression for $(\star\xi)_\top$ is incorrect - this form is going to have different expressions on different faces of the cube. In particular I think you used the original coordinate system with the expressions given on page 70 on that paper, while you needed to adapt your coordinate system separately for each face. Intuitively, the $\top$ operation is ...


2

What you need is pag. 43: understand notation in Lemma 4.1. Remark 4.2.; in particular formula 4.15. Formula 4.15 is interesting because for $u\in Y^N_n$ the r.h.s. vanishes (by definition of $Y^N_n$ $u=1$ on $\partial \Omega^N_n$). However, in Prop. 4.1. you need to consider non trivial finite energy elements in $X^3_n:=H(\mathbb T^3_n)$, though: so ...


2

The (Zariski closure of the) image of your map is studied in the following paper of Bruns and Conca: http://arxiv.org/abs/0705.3399 They stratify this closure by two numerical invariants rank and "small rank" and show that they correspond to $GL(V) \times GL(W)$ orbits. I never read the paper carefully, so I don't know if it will be of any use, but ...


2

I think you might just be forgetting that $f$ is $A$-linear. That makes $af(d)=f(ad)$ and $f(b)c=f(bc)$ and $f(b)f(d)=bdf(1)^2=0$. An easy visualization of the operation on $A\oplus A$ that you described is as the subring of the $2\times2$ upper triangular matrices over $A$ with the elements $$ \left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid ...


1

A toy example. Given an orthonormal basis $e_1,\dots,e_n$ for the vector space $V$, then $\nu=e_1\wedge\dots\wedge e_n$ and $$*(e_1\wedge e_2)=e_3\wedge\dots\wedge e_n\in \bigwedge^{n-2}V, $$ with $*e_1=e_2\wedge\dots\wedge e_n$ and $*e_2=-e_1\wedge e_3\wedge\dots\wedge e_n$ both in $\bigwedge^{n-1}V$. Introducing the insertion operator $i_\bullet: ...


1

First, let's settle the issue of the empty tensor product. Consider the identity map $id: W\to W$ for an arbitrary $R$-module $W$. This is an $R$-balanced map out of the collection $\{W\}\cup \emptyset$ and so descends to a map $W\otimes E\to W$. But we know that $(W,id)$ satisfies the universal property of the tensor product of $W$, so $W\cong W\otimes E$ ...


1

Gustavo, Are you sure this is all correct? The right approach is to use the fact that Lie derivative is a derivation and compute $L_S(X\wedge Y)$, etc. But I don't get your answers, and I think the hypotheses are mutually contradictory. In particular, the only way $[S,X]=mX$ can hold is to have $m=p$ and $X_2=0$. What is going on? :)


1

Tensor product is not exact, it is only right exact. But the tensor product does take split exact sequences to split exact sequences, i.e. tensoring $$M \to M \oplus N \to N$$ with $T$ gives $$T \otimes M \to (T \otimes M) \oplus (T \otimes N) \to T \otimes N$$ That means that if you have the inclusion of a direct summand $M \hookrightarrow E$ then ...


1

Here's one approach using the fact that projective modules are direct summands of free modules. Suppose $P\oplus Q\cong R^k$. Then $$R^{k^n}=(P\oplus Q)^{\otimes n}=P^{\otimes n}\oplus\cdots ,$$ so $P^{\otimes n}$ is projective. But then notice that $\Lambda^n(P)$ and $\mathrm{Sym}^n(P)$ are both direct summands of $P^{\otimes n}$.



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