# Tag Info

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

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

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., ...

6

Lets assume that $\dim V=m$ and $\dim W=n$ with $m\geq 2$ and $n\geq2$. Suppose that $\{v_1,v_2\}$ are linearly independent in $V$ and that $\{w_1,w_2\}$ are linearly independent in $W$. Seeking a contradiction, suppose that $$v_1\otimes w_1+v_2\otimes w_2=v\otimes w\tag{1}$$ Then extend $\{v_1,v_2\}$ to a basis $\{v_1,v_2,v_3,\dotsc,v_m\}$ of $V$ and ...

5

In one realization of forms, on a vector space $V$ over $\mathbf{R}$, a one-form is a linear map $V \to \mathbf{R}$. A two-form is an antisymmetric bilinear map $V \times V \to \mathbf{R}$. A three-form is an antisymmetric trilinear map $V \times V \times V \to \mathbf{R}$. And so forth. In case these words are unfamiliar, "trilinear", for example, means ...

4

Let $T(v_i) = \sum_j a_{ji} \cdot v_j$, so that $(a_{ji})$ is the matrix of $T$ w.r.t. to $(v_1,\dotsc,v_n)$. The wedges $v_{i_1} \wedge \dotsc \wedge v_{i_k}$ with $i_1<\dotsc<i_k$ form a basis of $\Lambda^k(V)$. We have: $$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} ... 4 It is necessarily possible to write all tensors as a simple tensor when either V or W has dimension 1 (or 0). I will leave this to you to confirm, the proof is fairly trivial. On the other hand, suppose that V and W have dimensions 2 or greater, and that we are given a basis of vectors v_i and w_i for the respective spaces. Then I am ... 4 I think the generalized complex number systems can simply be defined as rings of the form \mathbb R[X]/(f(X)) with f(X) a real quadratic polynomial and i is a notation of \overline X. For the meaning of this symbols maybe you can refer to wiki. 4 HINT: The image of A is a level surface of the function f(x,y,z)=z-F(x,y). 3 The statement in the article has to be taken with a huge grain of salt. Even for 1-forms, there's no consistent way to interpret all wedge products of 1-forms as intersections. If you restrict attention to nonvanishing decomposable forms (ones that can be written locally as wedge products of 1-forms), and add a few extra restrictions, then there is ... 3 At a non-critical point x of f, the kernel of the 1-form \text{d}f is the tangent space to the surface X := f^{-1}(f(x)) at x; similarly, for a non-critical point x of g, \text{d}g_x has the tangent space of Y := g^{-1}(g(x)) at x as its kernel. Putting both together, at non-critical points x for both f and g the radical of the ... 3 Like GFR, I also have no interest in an extended discussion. This post is meant to be an extended comment for the sake of adding perspective. Personally, I like to define the exterior derivative d in a coordinate-free, invariant way: Fact/Def: Let M be a smooth manifold (possibly with boundary). Then there are unique operators d \colon ... 3 The first map is just the induced map on the 2-graded component of the exterior algebra on K^3, namely \Lambda^2(K^3). The second is the map on the 2 tensors in the tensor algebra, i.e. elements of K^3\otimes K^3. On the exterior product we have the diagonal action, so determine the action on the basis in which your matrix is written. If the ... 3 (i) Your understanding is correct. (ii) The significance of the ordering is mainly bookkeeping; in particular they provide an easy way to start from a basis e_1,\dots,e_n for the vector space V and extend it to a basis for the spaces of p-vectors and the exterior algebra. (iii) An algebra is a vector space with a product (that satisfies some rules). ... 3 The answer to this question turned out to be rather uninteresting. There was an error on one of the Wikipedia articles which I had consulted, and this lead to my confusion. By the way, having done a bit more reading, I think that using the Plücker relations the way that I did, i.e. to decide when an element \alpha \in \Lambda^2(X), X a ... 3 Now apply the magic formula again:$$ L_X(\eta)=i_Xd\eta+di_X\eta $$where \eta=i_Yd\omega+di_Y\omega and get$$ L_X(L_Y\omega)=i_Xd(i_Yd\omega+di_Y\omega)+di_X(i_Yd\omega+di_Y\omega) $$use that d is linear$$ i_Xd(i_Yd\omega)+i_X(di_Y\omega)+di_X(i_Yd\omega)+di_X(di_Y\omega) $$Subtract L_Y L_X\omega and compare with$$ ...

3

To elaborate on Zhen Lin's comment: either viewpoint of a bilinear form is okay due to the universal property of the tensor product. Given a vector space $V$ with field of scalars $k$, I've usually seen a bilinear form defined as a bilinear map $B : V \times V \to k$, so a function that eats two vectors as in your first definition. However, by the ...

3

In general, let $V$ be a real or complex vector bundle of dimension $n$. What can we say about $V$ if it admits $k$ independent nonvanishing global sections? This is equivalent to admitting a splitting $V \cong W \oplus \mathbb{R}^k$ resp. $V \cong W \oplus \mathbb{C}^k$, and so it implies some conditions on characteristic classes. If $V$ is real, then ...

3

Since $$X\times Y=(X^2Y^3-X^3Y^2)e_1+(X^3Y^1-X^1Y^3)e_2+(X^1Y^2-X^2Y^1)e_3$$ then $$(X\times Y)^i=\eta_{ijk}X^jY^k,$$ is the correct formula for components. In the other hand if we agree the volume form be $e^{*1}\wedge e^{*2}\wedge e^{*3}$ and defined by $$e^{*1}\wedge e^{*2}\wedge e^{*3}=\sum_{\sigma\in S_3}(-1)^{\sigma} e^{*\sigma(1)}\otimes ... 3 I'll write out an answer for a form of three variables. This might yield some intuition. Thing to notice is that dx^i\wedge dx^i=0, and that dx^i\wedge dx^j=-dx^j\wedge dx^i, and f_{xy}=f_{yx}. Thus, because dx^i\wedge dx^i=0$$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz $$... 2 That's correct! We say a form \omega is closed if d\omega = 0, and we say that \omega is exact if \omega = d\eta for some form \eta. Your remark says, in this terminology, that every exact form is closed. However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare ... 2 The second definition is just wrong in characteristic 2. Any author who uses it is either implicitly ignoring the characteristic 2 case or is being imprecise. You find similar problems with the definition of Clifford algebras; there are again two variants and one of them is wrong in characteristic 2. 2 If \omega and \tau are closed forms (which you have not used), then$$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$hence \omega \wedge \tau is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of \omega and \tau. Let us suppose that \omega = ... 2 It should be stated in the problem but I think you are right about v_i = \pi (u_i). Hint: observe that \frac{1}{2}((u_1 + u_2) \otimes (u_1 + u_2))= \frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes u_1) + \frac{1}{2}(u_1 \otimes u_1 + u_2 \otimes u_2) is projected by \pi to 0 on one hand (projection of LHS) and to \frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes ... 2 Ref. 1 writes on p.10:$$ \Lambda^k T^{\ast}_{\mathbb{C}} M ~=~ \bigoplus_{j=0}^k \Lambda^{j,k-j} M,\tag{1.11}$$where we defined$$\Lambda^{p,q} M ~:=~ \Lambda^pT^{*(1,0)}M\otimes\Lambda^{q}T^{*(0,1)}M.\tag{1.11b}$$Here M is a 2n-dimensional real manifold with a complex structure J; the symbol \otimes denotes the standard ... 2 Here's a hint on your third thought. How does one differentiate a determinant? One row/ column at a time, holding the others fixed. Let V = [v_1,v_2,\cdots, v_n] where v_i' are columns. Then, look at the matrix e^{tA}V. Can you differentiate this function at t=0 in two ways that correspond to the two sides? 2 This picture for 1-forms is closely related to picturing 0-forms -- that is, ordinary functions -- in terms of their level curves; it's essentially the infinitesimal version of a contour plot. When following a curve along a contour plot, we can work out how much a function changes by counting how many contour lines it crosses over, and in which ... 2 There's a mistake in the last line in the calculation of i_{\hat{X}_t}\beta. The t's in the numerators should be in the denominators:$$i_{\hat{X}_t}\beta = \frac{-zx}{\color{red}{t}r^3}\, dy + \frac{zy}{\color{red}{t}r^3}\, dx = \frac{z}{\color{red}{t}r^3}(y\, dx - x\, dy).$$Then$$\Phi_t^*i_{\hat{X}_t}\beta = \frac{z}{(x^2 + y^2 + t^2 ...

2

A $k$-vector $w \in \bigwedge^kV$ is $m$-decomposable if there is a linearly independent set $\{e_1, \dots, e_m\}$ of $V$ and $\alpha \in \bigwedge^{k-m}V$ such that $w = e_1\wedge\dots\wedge e_m\wedge\alpha$; note that $k$-decomposable is what is normally called decomposable. Furthermore, $w$ is strictly $m$-decomposable if it is $m$-decomposable but not ...

2

First of all, note that in the formula you give us, $$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n},$$ there is a summation involved! In fact, there are $n-p$ sums, since the indices are being contracted. I won't use Einstein's convention in ...

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