Wikipedia:Reference desk/Archives/Mathematics/2010 December 29

= December 29 =

Inverse of an infinite tridiagonal matrix
Is there a particular condition for when we can define the inverse of an 'infinite' tridiagonal matrix? I'm not sure if the definition even really makes sense, but I want to look at a matrix over a basis over $$\mathbb{Z}$$, so infinite in 2 directions, and I want the matrix ('M', say) to satisfy $$M(e_i)=e_{i+1} \forall i \in \mathbb{Z}$$. Obviously because of countability we could reformulate the problem so that we only need to look at a basis over $$\mathbb{N}$$, but the form of the matrix (can we call it a matrix, if it's infinite dimensional?) wouldn't be as nice. Anyway, my problem involves looking at whether or not $$M-\lambda I$$ is invertible, the matrix in question having only $$-\lambda$$s on the diagonal and +1s on the subdiagonal. Is there going to be any specific way to tell when it's singular? I am aware for finite tridiagonal matrices we have only got to solve a set of simultaneous equations to find a solution, but I'm not sure when the infinite dimensional case is or isn't soluble. Ta all :) Spalton232 (talk) 04:08, 29 December 2010 (UTC)


 * Matrix threory is not likely to be helpful to you in this case; too many of its fundamental results depend on induction from one of the corners, which cannot even get started in the doubly-infinite case. You'd be better off with abstract linear algebra, simply looking at the linear transformations and ask if they are invertible. And that can depend not only on the coefficients, but also which vector spaces you're working with. For example, do you allow vectors with infinitely many nonzero entries?
 * In your particular problem, I think your transformation will not be invertible for any nonzero $$\lambda$$. If your vectors must have finitely many entries, then there's no vector that maps to $$e_i$$ for any $$i$$ (because the difference between the highest and lowest significant index always increases as you apply the map). On the other hand, if you allow infinitely many nonzero entries, then you can find nonzero input vectors that map to 0. –Henning Makholm (talk) 16:26, 1 January 2011 (UTC)

so what's a fractal dimension anyway?
and why is it a "dimension". Could you give some examples?

If you have an L shaped tetris block that is 1 unit deep like this (the numbers are how many units deep): [1] [1] [1][1] you can turn it along three dimensions getting either from the original to, turning clockwise in the first dimension (the plane of your screen) [1][1][1] [1] or you can get from the original to, turning about the plane of the table your screen is sitting on: [1] [1] [2] or you can get from the original to, turning about the plane cast by your eyes if you look up and down the exact center of your monitor, so that if you had laser vision you would cut it in two vertically: [3][1] fine. Great. Now please give me an example of doing the same thing with fractal dimensions. You can use any fractal you want, though of course you might have to inline some pictures instead of ascii art! Up to you... 88.182.221.18 (talk) 14:54, 29 December 2010 (UTC)


 * Fractal dimensions are not based on how many ways you can turn something. See Fractal dimension. PrimeHunter (talk) 14:59, 29 December 2010 (UTC)


 * For what it's worth, ordinary dimension is not based on the number of ways you can turn something either. A better explanation for why your Tetris-piece example shows that (three-dimensional) space has dimension 3 is that it is possible to have three mutually perpendicular lines in that space (the three axes you're rotating around)—the rotation itself has nothing to do with it. —Bkell (talk) 15:08, 29 December 2010 (UTC)


 * The last ASCII-art image seems wrong to me. IMHO it should be [3][1] --CiaPan (talk) 15:12, 29 December 2010 (UTC)

Sorry, fixed it!! Now, could you please do the same three operations on the following self-similar fractal, based on the cantor set: Step 1: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Step 2: aaaaaaaaaaaaaaaaaaaaaaaaaaa                          aaaaaaaaaaaaaaaaaaaaaaaaaaa Step 3: aaaaaaaaa        aaaaaaaaa                           aaaaaaaaa         aaaaaaaaa Step 4: aaa  aaa         aaa   aaa                           aaa   aaa         aaa   aaa

So, for me, it would be obvious to call the white-space dimensions, like this: Step 1: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Step 2: aaaaaaaaaaaaaaaaaaaaaaaaaaa111111111111111111111111111aaaaaaaaaaaaaaaaaaaaaaaaaaa Step 3: aaaaaaaaa222222222aaaaaaaaa111111111111111111111111111aaaaaaaaa222222222aaaaaaaaa Step 4: aaa333aaa222222222aaa333aaa111111111111111111111111111aaa333aaa222222222aaa333aaa

So it seems, like if I have the following fractal diagram: aaa333aaa222222222aaa333aaa111111111111111111111111111aaa333aaa222222222aaa333aaa

I should be able to "rotate" the 1, 2, and 3's. What would I get? Wouldn't the possible "orientations" be: aaa333aaa222222222aaa333aaa111111111111111111111111111aaa333aaa222222222aaa333aaa to aaa222aaa333333333aaa222aaa111111111111111111111111111aaa222aaa333333333aaa222aaa or to aaa333aaa111111111aaa333aaa222222222222222222222222222aaa333aaa111111111aaa333aaa or to aaa111aaa333333333aaa111aaa222222222222222222222222222aaa111aaa333333333aaa111aaa or to aaa222aaa111111111aaa222aaa333333333333333333333333333aaa222aaa111111111aaa222aaa or to aaa111aaa222222222aaa111aaa333333333333333333333333333aaa111aaa222222222aaa111aaa

It seems to me, like given any simple fractal: http://www.google.fr/images?q=simple+fractal Like I should just be able to "grab it" and rotate the fractal about fractal dimensions in much the same way as I described above!!

Am I just going crazy, or does what I say even make ANY sense whatsoever??? Thanks!!! 88.182.221.18 (talk) 17:55, 29 December 2010 (UTC)


 * I don't think you understand what dimension means, let alone fractal dimension. The concept of "dimension" has nothing to do with rotations. "Rotating a fractal about fractal dimensions" is completely nonsensical. —Bkell (talk) 18:27, 29 December 2010 (UTC)


 * you're right, a two-dimensional paint program could not possibly refer to something more limited when it says "rotate", then you refer to when you pick up an object in front of you and rotate it to orient in a different direction. Obviously, and as you correctly write, dimensions have nothing to do with rotation.  The only thing you have to specify when rotating an object is how many degrees you're doing the rotation by: dimensions have absolutely nothing to do with the rotation.  There is, after all, only one way to rotate a playing die clockwise by 180 degrees, just as there is only one way to rotate an object clockwise by 180 degrees in Microsoft Paint. 87.91.6.33 (talk) 19:36, 29 December 2010 (UTC)  87.91.6.33 (talk) 19:34, 29 December 2010 (UTC)
 * No, to rotate an object in 3D you need to specify an axis of rotation. In 4D you need to specify one or two planes of rotation. In general, a rotation is translation plus an orthogonal linear transformation of determinant 1. -- Meni Rosenfeld (talk) 20:26, 29 December 2010 (UTC)
 * Thanks Meni. Obviously what I (IP has changed) meant by "plane" in 3D rotation can be interpretered as the appropriate axis instead.  Could you please answer my question about Fractals?  I can give you many further examples if it would help you.  But the basic idea is this: the biggest dimension (level1) branches off into 5 smaller versions of the big picture (level2), those branch off into 5 (level3), those branch off into 5 level4's etc.  So, you grab one level2 branch: there are at least TWO ways of promoting it to level1.  Okay, it's level1 now, but there had been two dimensions hanging off of it: level3 and level4.  You could make level3 be now level4, and level4 stay level 4.  You could make level3 be level4 and level4 be level 3.  You could make what had been level3 and level4 both be now level3.  Etc.  You understand.  So, I want to learn about this kind of rotation using Fractal dimensions, Meni Rosenfeld.  Thank you for your help.  Maybe looking at a Fractal picture like this: http://api.ning.com/files/c-kkgOyXZLgfrSt8QT3WaqogZ8vBLYCqKC1At2ZLAYMrjg7bV1lcMX0N9h-uHkZImWby0qQFa44IIIrXwFXZ34cqt7hU*7w*/fractal_plant_image_s1.png can help you think while you look at what I wrote.  87.91.6.33 (talk) 22:08, 29 December 2010 (UTC)
 * What Meni meant there was you can actually rotate a 4D object in two different ways at different speeds at the same time. Also if constructing a fractal by making smaller copies the stages are not dimensions. Fractal dimension describes the idea, have you gone through why for instance the dimension of the Cantor set is log32? The idea you have of moving a fractal around is the idea of getting the symmetry group (mathematics) of the shape. Rotations are elements of a group so it is part of the same idea. Dmcq (talk) 23:56, 29 December 2010 (UTC)


 * In the real world the concept of a rotating a rigid body does not apply unless it is always rotating. If you rotated a sphere the equator would shrink and crack if it was rigid because of relativity, it only works because in our world things aren't rigid. Dmcq (talk) 19:48, 29 December 2010 (UTC)

Irreducible quartic polynomials with Galois group A4
Hello everyone,

I could use some help getting started with this problem - I have a lecturer who has given a thousand different lemmas and theorems and corollaries and no clue as to what to use when. Unfortunately, whilst I have a very effective toolbox for galois theory, I can't seem to find the right screwdriver. The problem is as follows:

Let P be an irreducible quartic polynomial over K with $$ char(K)\neq 2$$, whose Galois group is $$A_4$$. Show that its splitting field can be written in the form $$L(\sqrt{a},\sqrt{b})$$, where L/K is a Galois cubic extension and a, b ∈ L.

It's probably not a hard problem, but I'm struggling to get going - what should I do? Thankyou very much! Otherlobby17 (talk) 19:52, 29 December 2010 (UTC)
 * The crucial result is the appropriately-named fundamental theorem of Galois theory. The required field L corresponds to the subgroup V of $$A_4$$. L/K is Galois because V is normal, and cubic because V has index 3. The full splitting field has Galois group V as an extension of L, and looking at the subgroups of V (and using the quadratic formula) shows that it's of the given form. Feel free to ask again if you have trouble filling in any of the details. Algebraist 20:10, 29 December 2010 (UTC)

Homology and Cohomology
Let M be a smooth manifold. Let &Omega;k(M) denote the space of differential k-forms on M, &Gamma;k(M) the space of closed differential k-forms on M and &Lambda;k(M) the space of exact differential k-forms on M. the spaces &Omega;k(M), &Gamma;k(M) and &Lambda;k(M) are very useful, for example the k-th de Rham cohomology of M is given by the quotient space &Gamma;k(M)/&Lambda;k(M). But what is the homology and cohomology of the spaces &Omega;k(M), &Gamma;k(M) and &Lambda;k(M) themselves? — Fly by Night  ( talk )  22:20, 29 December 2010 (UTC)
 * You can only ask for the cohomology of something that forms a chain complex.
 * With &Omega;k(M), you have an exterior derivative d to &Omega;k+1(M), with d2 = 0. The cohomology is then given by the sequence of groups ker(d)/im(d), where it is understood that we take the appropriate versions of d (in the appropriate indices).
 * If you look only at closed forms, then you are looking at a chain complex, where the maps d are just the zero maps, by definition. Hence the cohomology of that complex is itself. Same for the exact forms. It's not something you're going to want to work with, as these spaces are going to be infinite dimensional most of the time. --SamTalk 16:57, 30 December 2010 (UTC)
 * Or did you mean something else when you asked for the cohomology of the "spaces themselves"? If you want to look at the de Rham cohomology of those treated as spaces, well, they are just vector spaces. The Rham cohomology should just be $$H^0(V,\mathbb{R}) = \mathbb{R}$$ and 0 in higher dimensions, by the Poincaré Lemma (there are slight subtleties associated to the fact that these spaces are infinite dimensional, but in any case you can consider the corresponding Čech cohomology and treat it as giving you the right answer). --SamTalk 17:04, 30 December 2010 (UTC)
 * Thanks for your comments. What I wanted was the homology and cohomology of the spaces &Omega;k(M), &Gamma;k(M) and &Lambda;k(M) themselves. So the underlying manifolds are &Omega;k(M), &Gamma;k(M) and &Lambda;k(M) themselves. In terms of the de Rham cohomology, we have a cochain complexes given by
 * $$0 \longrightarrow \Omega^0\left(\Omega^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^1\left(\Omega^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^2\left(\Omega^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^3\left(\Omega^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \cdots . $$
 * $$0 \longrightarrow \Omega^0\left(\Gamma^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^1\left(\Gamma^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^2\left(\Gamma^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^3\left(\Gamma^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \cdots . $$
 * $$0 \longrightarrow \Omega^0\left(\Lambda^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^1\left(\Lambda^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^2\left(\Lambda^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \Omega^3\left(\Lambda^k(M)\right) \stackrel{\text{d}}{\longrightarrow} \cdots . $$
 * I had thought that the space of differential k-forms is retractable and so, by the Poincaré lemma, must have trivial cohomology groups. Then by Poincaré duality, the homology groups must be trivial. But I wasn't sure if the infinite dimensionality caused problems. As for closed and exact, I guess you're right: they're vector spaces so they must have trivial co/homology groups. 'What about the relative homology groups over some coefficient ring R'':
 * $$ H_i(\Omega^k(M),\Gamma^k(M);R) \ \text{ and } \ H_i(\Omega^k(M),\Lambda^k(M);R) $$
 * Or '''What about the homology groups
 * $$ H_i(\Omega^k(M)/\Gamma^k(M);R) \ \text{ and } \ H_i(\Omega^k(M)/\Lambda^k(M);R) $$
 * I guess that in this last pair, we have vector spaces and trivial topologies yet again. — Fly by Night  ( talk )  20:16, 30 December 2010 (UTC)