5

Let $S^1$ be the unit sphere $x_1^2+x_2^2=1$ in $\mathbb{R}^2$ and let $X=S^1\times S^1\in\mathbb{R}^4$ with defining equations $f_1=x_1^2+x_2^2-1=0, f_2=x_3^2+x_4^2-1=0$. Show that the vector field $$w=x_1\frac\partial{\partial x_2}-x_2\frac\partial{\partial x_1}+\lambda\left(x_4\frac\partial{\partial x_3}-x_3\frac\partial{\partial x_4}\right)$$ ($\lambda\in\mathbb{R}$) is tangent to $X$ and hence defines by restriction a vector field $v$ on $X$.

What does "defines by restriction" mean here? $w$ is already a vector field. Does it just mean that the vector field $v$ is the same one but restricted from $\mathbb{R}^4$ to $X$? Then we don't need this vector field to be tangent to $X$ in order to define the restriction, do we? (Note the use of the word "hence" in the problem statement.)

EDIT: This is the definition of a vector field I'm using: A vector field on an open set $U\in\mathbb{R}^n$ is a function $v$ which assigns to each point $p\in U$ a vector $v(p)$ in the tangent space $T_p\mathbb{R}^n$. The tangent space $T_p\mathbb{R}^n$ is just the set of pairs of the form $(p,v)$, where $v\in\mathbb{R}^n$. From this definition, it would seem we always obtain a vector field on $X$, don't we?

EDIT 2: I think I see where I'm going wrong now. The above definition works only on an open set $U$. Here $X$ is not an open set, so we need the vector $v(p)$ to be in the tangent space $T_pX$.

PJ Miller
  • 8,193
  • 2
    w is a map from $\mathbb{R}^4$ to its tangent bundle. You are restricting this map to the torus, $T^2$. This factors through the inclusion of the $TT^2$ (the tangent bundle of the torus) into $T\mathbb{R}^4$ (the tangent bundle of $\mathbb{R}^4$) and so defines a vector field on $T^2$. – Tim kinsella Nov 09 '13 at 06:51
  • @Timkinsella I'm sorry, but I'm not familiar with tangent bundles or torus yet. Could you please perhaps explain it in more elementary terms? Thanks. – PJ Miller Nov 09 '13 at 06:55
  • Sure. What is your definition of a vector field? The torus is just your $S^1 \times S^1$ – Tim kinsella Nov 09 '13 at 06:57
  • @Timkinsella Sure. A vector field on $U\subseteq \mathbb{R}^n$ is a function $v$ which assigns to each point $p\in U$ a vector $v(p)$ in the tangent space $T_p\mathbb{R}^n$. The tangent space $T_p\mathbb{R}^n$ is just the set of pairs of the form $(p,v)$, where $v\in\mathbb{R}^n$. – PJ Miller Nov 09 '13 at 16:44
  • I am confused why, in the problem statement, it seems like we need the vector field to be tangent to $X$ in order to be able to define this restriction. Can't we always define the restriction, no matter whether the vector field is tangent to $X$ or not? – PJ Miller Nov 09 '13 at 17:14
  • 1
    I think the problem is that you're working from the wrong definition. Yours is only reasonable when $U$ is an open subset of $\Bbb{R}^n$. For more general submanifolds of $\Bbb{R}^n$, there should be some tangency condition built into the definition. (Vector fields on $X$ are supposed to be part of the intrinsic geometry of $X$, so they shouldn't depend on how you choose to embed $X$ in Euclidean space; if you left off the tangency condition, different embeddings would give different sets of vector fields.) – Micah Nov 09 '13 at 17:30
  • @Micah I'm sorry.. in my definition $U$ is indeed an open subset. Now would you please reconsider your answer? – PJ Miller Nov 09 '13 at 19:10

1 Answers1

5

Here is a general answer which, unfortunately, I couldn't locate in the standard textbooks.

Let $L \subset \mathbb R^n$ be a submanifold of dimension $n-k $ defined by the equations $f_i=0$, where the $f_i\; (i=1,...,k)$ define a submersion $\mathbb R^n\to \mathbb R^k$.
A vector field $w=\sum v_j\frac {\partial}{\partial x_j}$ is tangent to $M$ at $m\in M$ iff $$w(f_i)(m)=\sum_{j=1}^n v_j\frac {\partial f_i(m)}{\partial x_j}=0$$ for all $i=1,...,k$.

Applying this to your problem immediately solves it : check that $$w(f_1)=x_1\cdot 2x_2-x_2\cdot 2x_1=0, w(f_2)=\lambda [x_4\cdot 2x_3-x_3\cdot 2x_4=0]$$

Georges Elencwajg
  • 150,790
  • I think it is an unfortunate state of affairs that many students could give you in detail abstract definitions of tangent bundless, vector fields, differential forms and their derivation from multilinear algebra and still be unable to solve a little exercise like the present one. In my opinion the pendulum has swung too far in university teaching in its move from calculations to abstract concepts. – Georges Elencwajg Nov 09 '13 at 10:26
  • Thank you for your answer, Georges. I appreciate your comment, but at the same time my problem is that I am confused why, in the problem statement, it seems like we need the vector field to be tangent to $X$ in order to be able to define this restriction. Can't we always define the restriction, no matter whether the vector field is tangent to $X$ or not? – PJ Miller Nov 09 '13 at 17:13
  • Yes, you can restrict a vector field to $X$ but what you obtain is not a vector field on $X$ but a section of the vector bundle $T_{\mathbb R^4}|X $ which is much bigger than $T_X$: we have $T_X \subsetneq T_{\mathbb R^4}|X $ and these vector bundles on $X$ have rank respectively $2$ and $4$. – Georges Elencwajg Nov 09 '13 at 18:27
  • This is the definition of a vector field I'm using: A vector field on an open set $U\in\mathbb{R}^n$ is a function $v$ which assigns to each point $p\in U$ a vector $v(p)$ in the tangent space $T_p\mathbb{R}^n$. The tangent space $T_p\mathbb{R}^n$ is just the set of pairs of the form $(p,v)$, where $v\in\mathbb{R}^n$. From this definition, it would seem we always obtain a vector field on $X$, don't we? – PJ Miller Nov 09 '13 at 19:14
  • That definition is useless here: it does not cover the case of a closed submanifold, which is the sort of manifold that $X=S^1\times S^1\subset \mathbb R^4$ is. – Georges Elencwajg Nov 09 '13 at 19:19
  • Yes, I think I see where I'm going wrong now. If I have another question I'll start a new question instead. Thanks Georges! – PJ Miller Nov 09 '13 at 19:21
  • I'm happy this point is settled. I will certainly be happy to read your new question. – Georges Elencwajg Nov 09 '13 at 19:26
  • @GeorgesElencwajg , I'm not sure its fair to make such sweeping judgments about pedagogy based on the commenters' inability to guess the OP's nonstandard definition of a vector field. – Tim kinsella Nov 09 '13 at 20:20
  • @Tim: My comment is not based on any commenter's inability but on decades of teaching and hearing from students: "I understand the theory perfectly, but can't do the exercises". Since these were my students, my criticism applies to me too. My comment is also based on the 966 questions I answered here and the many more I read without answering. And, by the way, to whom do you think I'm being unfair? – Georges Elencwajg Nov 09 '13 at 20:42
  • @GeorgesElencwajg : Oh I misunderstood you then. In fact I very much agree that students should do more computations. I do, however, find it ironic that in this post, the only obstacle for the OP lay in recognizing the need for a more abstract definition of the notion of a vector field, not in carrying out a computation. – Tim kinsella Nov 09 '13 at 21:09
  • 1
    Dear @Tim, what also motivated my comment is the fact that the criterion I use is not emphasized, nor even mentioned, in textbooks (as far as I know) and that's a pity. – Georges Elencwajg Nov 09 '13 at 21:13
  • @GeorgesElencwajg : I agree, and I'm happy to know it now! – Tim kinsella Nov 09 '13 at 21:14
  • @GeorgesElencwajg : I think what motivated my comment was a feeling that I should come to the defense of the abstractions that are so dear to me. I think that maybe a pendulum may not be the best metaphor, and that computation and praxis needn't come at the expense of abstraction and theory. In any case, this is a big topic, and this probably isn't the best place for a discussion :) – Tim kinsella Nov 09 '13 at 21:26
  • Dear @Tim: I also love big abstract machines (after all I am a 21st century algebraic geometer) but I think they should be counterbalanced by a deliberate decision to be an old-fashioned craftsman too. Since life is short and art long, these requirements aére difficult to fulfill simultaneously . But you are right, here is not the best place for a discussion... – Georges Elencwajg Nov 09 '13 at 21:36
  • Here is a new (30 minutes ago) illustration of what I mean: the OP has a splendid command of theoretical scheme theory but has difficulties with computations. – Georges Elencwajg Nov 09 '13 at 22:07
  • @Tim... and that was a few hours ago. – Georges Elencwajg Nov 09 '13 at 22:13