Timeline for Clifford algebra as an adjunction?
Current License: CC BY-SA 2.5
8 events
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Dec 3, 2009 at 21:36 | comment | added | Mariano Suárez-Álvarez | By the way, the fact that a left adjoint preserves coproducts of course is true, and one can construct coproducts in the category of Z_2 graded algebras (imitating the construction of the free product of groups) so you can make something out of this argument. | |
Dec 3, 2009 at 19:44 | comment | added | José Figueroa-O'Farrill | Yes, but there are useful ways of being wrong :) | |
Dec 3, 2009 at 18:57 | history | edited | Alicia Garcia-Raboso | CC BY-SA 2.5 |
added 66 characters in body
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Dec 3, 2009 at 18:55 | comment | added | Mariano Suárez-Álvarez | In any case, the Clifford algebra of an orthogonal direct sum of quadratic spaces is isomorphic to the twisted tensor product of the corresponding tensor algebras: this is the "super" tensor product, the one which introduces signs by the parity graduation of the Clifford algebras. | |
Dec 3, 2009 at 18:54 | comment | added | Alicia Garcia-Raboso | My bad: Clifford algebras are Z_2-graded, and hence the coproduct is the Z_2-graded tensor product. Proposition 1.5 on page 11 of Lawson and Michelsohn's "Spin Geometry" asserts that Cl indeed preserves coproducts. | |
Dec 3, 2009 at 18:53 | comment | added | José Figueroa-O'Farrill | Curiously, since Clifford algebras are $\mathbb{Z}_2$-graded, if you were to take the $\mathbb{Z}_2$-graded tensor product in your first displayed equation, then this would be a true statement. This perhaps suggestst that I have to consider the category of $\mathbb{Z}_2$-graded associative algebras. | |
Dec 3, 2009 at 18:42 | comment | added | Mariano Suárez-Álvarez | The coproduct in the category of associative algebras is not the tensor product: two maps $f:A\to C$ and $f:B\to C$ give a map $A\otimes B\to C$ only if the images of $f$ and of $g$ commute. | |
Dec 3, 2009 at 18:27 | history | answered | Alicia Garcia-Raboso | CC BY-SA 2.5 |