# Tag Info

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

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, ...

6

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 ...

6

$$v \otimes w + w \otimes v = (v+w) \otimes (v+w) - v\otimes v - w\otimes w$$ so the latter ideal contains the first one. For field of characteristic not 2, the first one contains the latter one. So in $char \neq 2$ cases they are the same. I believe that $v \wedge v = 0$ is a condition you always want, which cannot be deduced from your first definition. ...

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 ...

5

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' = ... 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 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 $$Thus if \gamma_1  and \gamma_2  are curves with same ... 4 The lateness of this answer gives a new meaning to the word "overdue". In any case, I noticed this question while browsing this website and I thought I'd answer it because that's what you do when you're faced with a question right? I'm not sure if this is what you're looking for; it's basically equivalent to your basis-dependent proof above and it's based ... 3 This is a variant of the polarization trick:$$ v \otimes w + w \otimes v = (v + w) \otimes (v + w) - v \otimes v - w \otimes w $$Let I be your ideal generated by all v \otimes w + w \otimes v and let J be the ideal generated by all w \otimes w. The polarization identity shows that I is contained in J. Of course, you can go the other way ... 3 Not sure if this is a rigorous explanation but an intuitive way to think of this: The idea of quotiening out by the ideal \{ v \otimes w+ w \otimes v | v,w \in W\} is that we impose the relation$$v \wedge w = - w\wedge vin the quotient where v \wedge w is the image of v \otimes w under the canonical map. Now what happens when we quotient out by ... 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 These definitions are the same. In fact, there is a third, more common definition that they are equivalent to. Namely, if we take a local coordinate system (x^{1}, \cdots, x^{n}), we get dx^1 , \cdots , dx^n as a basis for the cotangent space of a point in the chart. If we take a multi-index I = (i_{1}, \cdots, i_{k}), then we can consider a k-form ... 3 Green's Theorem states that, for a simply connected region D:\oint_{\partial D} (P dx + Q dy) = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right )$$Choose P=-y and Q=x to get for the area A(D)$$A(D) = \frac{1}{2} \oint_{\partial D} (-y \,dx + x \,dy)$$For an ellipse, x=a \cos{t}, y=b \sin{t}, and we ... 3 As you suspect the answer is yes, and you basically explain why it is so. If you are not convinced you could add the following details: your PDE a_1 f_x + a_2 f_y=0 is first order with the variable f the dependent variable while x,y,w,z are the independent variables. Standard coordinates in first jet space are x,y,w,z,f,f_x,f_y,f_w,f_z while the ... 3 We will use http://en.wikipedia.org/wiki/Exterior_algebra as a reference Here$$(f \wedge g) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) f(x_{\sigma (1)}, \dots, x_{\sigma(k)}) g( x_{\sigma(k+1)}, \dots, x_{\sigma(k+l)}) (af \wedge bg) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) a f(x_{\sigma (1)}, \dots, ...

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

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 ...

3

I will keep referring to the Leibniz-rule for exterior derivatives. That is axiom 3 here. The expression $\rho \wedge \mathrm{d}\rho = 0$ is only true for all one-forms $\rho$ in less than or equal to two dimensions. In dimension $\geq 3$, let $x,y,z$ be the first three of the coordinate functions, we can consider the one form $$\rho = x \wedge ... 2 If your elements commute with eachother, then there is no need for an ordering in the case of finite sums/products. In the non-commutative case things are more complicated. Anyhow, IMO there is no need for an ordering if the sum/product doesn't depend on the order. And this covers many non-commutative cases too. Otherwise, it is clear that one should ... 2 By definition, if V is a vector space (forget about it having a basis) over a field k, then the dual space V^* is the space of linear functionals f\colon V\to k. Thus the most natural thing to do if you have a linear functional f\in V^* and a vector v\in V, is to take f(v). This gives you the natural pairing V^*\times V\to k, (f,v)\mapsto ... 2 The following hint works if you suppose \beta to be closed. I'm not sure (though I doubt) that \eta \wedge \beta is in general exact if \eta is exact. Try to write out what it means for \eta to be exact, i.e. that \exists \omega such that \eta=d\omega. Then recall the formula:$$ d(\alpha \wedge \gamma)=d\alpha \wedge \gamma + (-1)^k \alpha ...

2

Maybe it would help to work with algebras rather than coalgebras: since everything is finite-dimensional, we can dualize. The dual to the coalgebra $E(x)$ is again the exterior algebra $E(x)$, and we can work with $E(x)$-modules rather than comodules. In this case, the goal is to show that $\mathrm{Ext}^{\bullet}_{E(x)}(k, k)$ is a polynomial algebra, ...

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

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

One problem with your approach is that your basis is too generic, which is inconvenient for computations. Try some special basis: e.g. Combine $a_1e_1 \wedge e_2 + a_2 e_1 \wedge e_3 + a_3 e_1\wedge e_4$ as $e_1 \wedge (a_1e_2 + a_2e_3 + a_3e_4)$. Then let $e_2' = a_1e_2 + a_2e_3 + a_3e_4$ we can replace the sum of first three terms as $e_1 \wedge e_2'$. ...

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

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\$, ...

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