Talk:Multicomplex number

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Mathematics rating: Stub Class Low Priority  Field: Algebra

Hello, 8OYscLu9. Your earlier remarks are fully justified in my eyes. The statement that the bicomplex numbers are a special case holds true, however, both articles use different reference material I believe. Well, and then I may have made a mistake altogether, who knows. While I was glad to create the multicomplex number stub from the reference given, the bicomplex number article uses a web reference with different definitions. In short, if instead of real number coefficients you use another multicomplex number as coefficients, you'll again end up with a multicomplex number. So, I find your "clean-up please" remark appropriate and would leave it out here ... until someone has the time to pick it up ... Thanks, Jens Koeplinger 14:57, 20 January 2007 (UTC)

[edit] It's a bit better now I think

I've taken Jheald's latest additions and grouped it a bit. I think the article is getting better now. Maybe some general properties or so, and it should be a nice article. Thanks, Jens Koeplinger 12:01, 1 April 2007 (UTC)

PS: But when direct sum and when outer product should be used to indicate isomorphism. Koeplinger 12:16, 1 April 2007 (UTC)
The two products \oplus and \otimes are very different things. If we take \Bbb{C}\otimes\Bbb{C}\otimes\Bbb{C} that means that each base of the algebra can be factored into a power of i1 times i2 times i3, so each number in \Bbb{C}\otimes\Bbb{C}\otimes\Bbb{C} is a linear combination over eight basis elements.
On the other hand, when we write \Bbb{MC}_8 = \Bbb{C}\oplus\Bbb{C}\oplus\Bbb{C}\oplus\Bbb{C} that means each of its 8 basis elements can be written as a particular vector of 4 complex numbers, with the product of two vectors defined by the product (?or should that be a product?) of only the corresponding components, taken pair by pair.
Are the two isomorphic? Can one write \Bbb{C}\otimes\Bbb{C}\otimes\Bbb{C} \sim \Bbb{C}\oplus\Bbb{C}\oplus\Bbb{C}\oplus\Bbb{C} ? I'm not sure. Can one find an element of the first space such that α8 = − 1, and none of the numbers from α...α4 can be expressed as linear combinations of their predecessors? I don't know; exp(i1π / 4)exp(i2π / 4)exp(i3π / 4) might be a candidate. Can one find eight different members u of the second space such that u2 = -1 ? ... Left as an excercise for the reader.  :-) Jheald 14:06, 1 April 2007 (UTC)
Ok. I've added the word "also" to the isomorphism that uses an outer product (diff) to indicate that this is shown with purpose (and not a mistake). For n=4 we have \Bbb{C}\otimes\Bbb{C} = \Bbb{C}\oplus\Bbb{C}, but this is a special case. Thanks for the details, Jens Koeplinger 14:31, 1 April 2007 (UTC)