Wikipedia:Reference desk/Archives/Mathematics/2011 March 8

= March 8 =

Points within a fixed hyperbolic distance of a geodesic in the unit disc B^2 and unit ball B^3
Hi everyone,

I've been asked to draw the set of points within a fixed hyperbolic distance of a geodesic in the unit disc $$\mathbb{D}$$ and unit ball B$$^3$$, but the trouble is I haven't the faintest clue what they look like. I was hoping someone might be able to link me to a picture perhaps, as I was unable to track one down myself. I'm not sure whether we're safe to assume ('WLOG') that the geodesic is e.g. the real axis in the unit disc, or whether that vastly alters the picture from the general case.

I can figure out the obvious things - e.g. distances become arbitrarily large at the edges of the disc/ball, so any set of points we do have is going to taper off to a point as we near the edges of the disc, and 'bulge' in some manner near the origin (or the point nearest the origin, if we're not taking the real axis wlog), but I don't know what the actual point set looks like outside of those two facts. I guess in the ball case things get more complicated, so any links you could suggest for a picture or any help you could give would be great. Thanks very much! Estrenostre (talk) 00:03, 8 March 2011 (UTC)


 * I think the Equidistant curve article answers your question at least in the two-dimensional case, assuming that by $$\mathbb{D}$$ you mean the Poincaré disk model. –Henning Makholm (talk) 01:00, 8 March 2011 (UTC)

Tensor Question
The tensor article says that a type (1,1)–tensor is a map V ⊗ V* → R for some vector space V. While Tensor (intrinsic definition) says that the type (1,1)–tensors are isomorphic in a natural way to the space of linear transformations from V to V. Can anyone explain why? — Fly by Night  ( talk )  13:45, 8 March 2011 (UTC)
 * There is a natural isomorphism between
 * $$L^2(V\times V^*, \mathbb{R}) \cong V\otimes V^*$$
 * so a (1,1) tensor is defined as an element of one or the other, depending on the author. There is a tensor-hom adjunction giving an isomorphism
 * $$Hom(V,V)\cong V\otimes V^*$$
 * - Sławomir Biały (talk) 14:02, 8 March 2011 (UTC)
 * Ermm... Errr... That's all a bit too general for me, to be honest. Could you give an explicit isomorphism, or at least explain how it might work? Thanks. — Fly by Night  ( talk )  14:21, 8 March 2011 (UTC)
 * The standard approach is to verify the isomorphisms in a basis, and show that they are basis-independent. Sławomir Biały  (talk) 14:37, 8 March 2011 (UTC)


 * I'll take a stab at a less rigorous, more "hand waving" answer. Pick a specific basis, and you can represent a (1,1)-tensor by a matrix (of course, the matrix has to transform in the "right way" if you change your basis). Multiply a row vector by the matrix and you get another row vector - so the matrix also represents a linear map from V to V. Multiply the matrix by a row vector on the left and a column vector on the right and you have a scalar - so the matrix also represents a bilinear map from V x V* to R. So (1,1) tensors, linear maps from V to V and bilinear maps from V x V* to R are just different ways of describing the same mathematical object. Gandalf61 (talk) 15:12, 8 March 2011 (UTC)
 * Excellent! Thanks Gandalf, much appreciated.


 * Hmm, this equivalence works only for finite-dimensional V, right? (Or to a settings where the maps are restricted such that $$V^{**}$$ is naturally isomorphic to V). There's always a natural bijection between (1,1)-tensors and linear transformations $$V\to V^{**}$$, but plain $$V \to V$$ may be smaller than that. –Henning Makholm (talk) 16:58, 8 March 2011 (UTC)
 * Could you explain what you mean in more detail, please? Gandalf's explanation doesn't pose any problems for my setting. The linear maps in question are transformations of tangent spaces to finite dimensional manifolds. — Fly by Night  ( talk )  17:34, 8 March 2011 (UTC)
 * Sorry. My comment was confusing and partly irrelevant. Since writing it I've almost convinced myself that the concept of "tensor of type (n,k)" is not actually used for general infinite-dimensional vector spaces, which moots the main point I thought I was making.
 * One source of confusion that I was not aware of is that there are two different definitions in play here. The one you quote from tensor says that an (1,1) tensor is a bilinear map V*×V &rarr; R; the one adopted by tensor (intrinsic definition) says that an (1,1) tensor is an element of the tensor product V&otimes;V*. These two definitions are equivalent when V is finite-dimensional, but not in general.
 * The argument I had in mind works from the V* × V &rarr; R defintion. Let T be a linear transformation V &rarr; V**. Then we can define &tau; : V* × V &rarr; R by &tau;(w,v) = T(v)(w). Conversely, the same equation defines T if we already know &tau;. In the finite-dimensional case, V and V** are naturally isomorphic, so V &rarr; V** is just the same as a linear transformation V &rarr; V.
 * In contrast, if the tensor is defined as an element of the tensor product V &otimes; V*, such an element is in general a formal sum $$(v_1,w_1)+...+(v_k,w_k)$$ of pairs of vectors and covectors, modulo a certain equivalence relation. Then we can easily get from the tensor to a linear transformation T:V&rarr;V by setting $$T(v) = w_1(v)\cdot v_1 + \cdots + w_k(v)\cdot v_k$$. On the other hand, getting from a general linear transformation T to a tensor is now only possible if V has a finite basis $$(b_1,...,b_n)$$, in which case T corresponds to the tensor represented by $$(T(b_1), \langle b_1,-\rangle)+\cdots+(T(b_n),\langle b_n,-\rangle)$$, where $$\langle b_i,-\rangle$$ is the element of V* that takes each v to its i'th component when expressed in the basis $$(b_n)_n$$. –Henning Makholm (talk) 20:36, 8 March 2011 (UTC)

You can find a detailed explanation of the equivalence of the different definitions (albeit for (0,2) tensors) in the article metric tensor. Sławomir Biały (talk) 00:38, 9 March 2011 (UTC)
 * Thanks, but I already understand metric tensors. It's the fact that type (p,q)−tensors can be isomorphic to other things, as you mentioned with the tensor-hom adjunction link, that confuses me. I'm only a humble differential geometer. — Fly by Night  ( talk )  01:28, 9 March 2011 (UTC)
 * Ok. Equivalence of the different definitions proceeds in the same way for p,q tensors as well.  I can't respond at length right now though. Sławomir Biały  (talk) 02:07, 9 March 2011 (UTC)