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Algebra isomorphisms

  1. Sep 7, 2010 #1
    can anyone tell me when one algebra is said to be isomorphic to another algebra?.

    I am interested in the (Clifford) algebras that are isomorphic to the quaternion-algebra.
    I know that, for example, the even subalgebra of [tex]\mathcal{C}l_{3,0}[/tex] is isomorphic to the quaternion algebra.
    However I am interested in finding other isomorphic algebras, and for that I need a rigourous definition of "algebra isomorphism".

  2. jcsd
  3. Sep 7, 2010 #2


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    If you view an algebra simlpy as a set (ring, module, whatever) A with a bilinear product [., .] then isomorphism of two algebras (A, [., .]) and (B, {., .}) means what you would expect, namely:
    there is an isomorphism [itex]\phi: A \to B[/itex] such that
    [tex]\phi([x, y]) = \{ \phi(x), \phi(y) \} [/tex]
    for every x and y in A.

    In terms of your example, if [itex]A = \mathcal{C}l_{3,0}[/itex] and [x, y] := xy (the Clifford product) and [itex]B = \mathbb{H}[/itex] (the quaternions) with {x, y} = xy (the usual quaternion multiplication) you can define the isomorphism by
    [tex]\phi(1) = 1, \phi(\hat e_1 \hat e_2) = i, \phi(\hat e_2 \hat e_3) = j, \phi(\hat e_3 \hat e_1) = k[/tex]
    and extend it linearly.
  4. Sep 7, 2010 #3
    Thanks a lot CompuChip,
    the answer was very clear.

    So, for the particular case of testing whether f is a isomorphism between two Clifford Algebras, we have to prove that it preserves the Clifford product, and it is of course linear (and bijective). True?
  5. Sep 7, 2010 #4


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    Yep, and actually that is the definition of an isomorphism in about any setting. It should be bijective (i.e. the two things that are "the same" can be mapped one-on-one) and it should preserve whatever operation you have between them (i.e. it doesn't matter if you apply the operation in one and then apply the isomorphism to the result, or if you first apply the isomorphism to both elements and then the operation of the other).

    If you ever get the chance to learn some category theory, you should take a look... there it is made very explicit that all isomorphisms are just bijections between objects (groups, algebras, vector spaces etc) which preserve "the" operation (resp. group multiplication, algebra operations, vector addition, etc)
  6. Sep 7, 2010 #5


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    And the inverse "preserves the operation".
  7. Sep 8, 2010 #6


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    In general category theory this is not true, since of course objects need not be sets (in which case arrows cannot be "bijective"). The general definition of isomorphism is an arrow which is both left- and right-invertible. In the concrete categories you mention (groups, algebras, vector spaces,...) this amounts to bijective arrows which preserve the structure and whose inverse does. And in some of those categories it turns out that the inverse automatically preserves the structure (but not always, e.g. in Top: a bijective continuous map need not be a homeomorphism).
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