# 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

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 ...

4

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 ...

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

There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively. In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.

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

As Darij remarked in his comment, if $R$ is not a ${\mathbb Q}$-algebra one cannot argue that $\bigwedge^n P$ and ${\mathfrak S}^n P$ are summands of $P^{\otimes n}$, thereby reducing the statement to showing that projectivity and flatness are stable under taking tensor powers. Instead, one might argue as follows: Freeness: If $F$ is a free $R$-module on ...

4

HINT: The image of $A$ is a level surface of the function $f(x,y,z)=z-F(x,y)$.

4

I think there is an importan point that has been overlooked in the above answers: The exterior derivative is the only linear natural operator in the list. This is explained with several variations the book by Kolar, Michor and Slovak cited in Yuri Viatkin's answer. The Lie derivative is also natural under general diffeomorphisms but only as a bilinear ...

3

They both may be right. If wedge product differs (Similar problem) and we set the definition of exterior derivative as $$d\omega=\sum_{I}d\omega_I\wedge dx^I,$$ then $d$ may differs as well (cause $\wedge$ appears). If we take axiomatic approach to exterior derivative, then one of axioms says ...

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 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

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 ...

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 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 $$... 3 The easiest way to answer your question is first bring it in an 'equivalent' form and then answer it. Of course you might still argue whether the the reversed form is really equivalent... The modified question is: Can we identify e_1 \wedge \cdots \wedge \hat{e}_i \wedge \cdots \wedge e_N with e_i^\perp? (So I only moved the \perp to the other side ... 3 If \omega and \tau are closed forms (which you have not used), then$$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$hence \omega \wedge \tau is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of \omega and \tau. Let us suppose that \omega = ... 3 Since$$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 ... 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 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

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

Because the similarity between symmetric tensor product and wedge product, I will discuss only the wedge product here. It is common to see both two definitions of wedge product in different textbooks. Given two differential forms $\alpha\in\bigwedge^p(V)$ and $\beta\in\bigwedge^q(V)$, we can define the wedge product as $\displaystyle ... 2 Comments to the question (v2): One can define left and right exterior derivatives $$d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L},$$ $$d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R},$$ $$d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C}$$ where$\omega, ...

2

We do, but for historical reasons they are called (linear) isometries.

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

In general, consider a one form $\beta = \sum_i \beta_i dx^i$ on $U\subset\mathbb R^m$. If $$\| \beta\|^2 = \beta_1^2 + \cdots + \beta_m^2 \neq 0$$ on $U$, then the $(n-1)$-form $$\alpha = \frac{1}{\|\beta\|^2} \sum_i (-1)^{i-1} \beta_i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^m$$ on $U$ safisfies \beta \wedge \alpha = dx^1 ...

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