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 Exercise [15.04] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [15.04]
My solution is attached.


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03 Feb 2011, 13:15

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [15.04]
Note that for higher-dimensional algebras that don't support division, the first problem I mention might possibly(*) be obviated by simply defining:

M = { w=Cz: C in X } U { z=0 }

At first glance this looks like M is not a well-defined manifold, since there's no coordinate patch that includes the "z=0" element. But we can simply create one. The important property here is that M be a manifold, with the (real) components of C forming one possible coordinate system on (most of) it.

Just define D = (1/|C|^2).C, where |.| is the regular euclidean vector norm (square root of sum of components squared).

Then the components of D also form a coordinate system on M, and the patches overlap everywhere except C=0 and D=0. We can then associate the point D=0 with the "z=0" element.

(*) There's an important caveat, however: we need "z=0" to be the limit of "w=Cz" as D "approaches zero" or C "approaches infinity" in magnitude; and we run into problems here if we convert w=Cz into a matrix equation, and the matrix C is degenerate. We need it to have a nonzero determinant for this limit to be valid. This is equivalent to saying that the algebra X has no divisors of zero (meaning no nonzero elements x, y such that xy=0)

In any case, there's still the second problem: we still need a way to prove that there's a unique C that satisfies w=Cz for any given (w,z) on B (z nonzero), because that's what's required to define the canonical projection from B to M. With division we could simply take C=w/z; without it we can still proceed by converting the linear relation between C and w into a matrix equation (this time with w and C as the vectors and z as the matrix part); a unique solution then exists provided the matrix z is invertible... once again this is equivalent to saying that the algebra X has no divisors of zero.

So, unless I've missed something, the key feature of the algebra X that's required for (a suitably modified version of) the argument to work properly appears to be: No divisors of zero.

This is a weaker condition than the existence of division: Division implies it; but not the other way around.


It's certainly interesting that the topological properties of spheres should be so closely interrelated with this feature of higher-dimensional algebras!


03 Feb 2011, 16:37
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