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Why the tensor product of graded algebras is defined with a commutation $\epsilon $ like this : $(a\otimes b)(c\otimes d)= \epsilon(ac\otimes bd)$ ? what is the usefulness of the commutator $\epsilon$ here ? why that tensor product is not defined simply by $(a\otimes b)(c\otimes d)= (ac\otimes bd)$ ?

Albert_T
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  • Where did you get this definition? I have never seen the tensor product of algebras defined with that element. I even double checked: https://en.wikipedia.org/wiki/Tensor_product_of_algebras Maybe for clarity you should add something like: "I have two $R$-algebras $A$ and $B$, etc..." – acyrl Sep 26 '12 at 09:49
  • see Bourbaki, Algebras chapters 1 to 4 – Albert_T Sep 26 '12 at 10:04
  • Sorry, I had missed the graded bit. – acyrl Sep 26 '12 at 11:16
  • Even taking into account superalgebras and the Koszul sign rule, I don't think $(a\otimes b)(c\otimes d)= \epsilon(ac\otimes bd)$ is anywhere near the correct formula. – darij grinberg Oct 08 '15 at 00:27

1 Answers1

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The use of the $\epsilon$ here can be to inherit structure from $A$ and $B$: If $A$ and $B$ are for example graded anticommutative algebras, and you want $A \otimes_R B$ to be graded anticommutative also, $(a \otimes b)(c \otimes d) = (ac \otimes bd)$ won't do: Say for example $a \in A_0$, $c \in A_1$, $b \in B_1$, $d \in B_0$. The you have $ac = ca$, $bd = db$ and hence \[ (a \otimes b)(c \otimes d) = (ac \otimes bd) = (ca \otimes db) = (c \otimes d)(a \otimes b) \] But if we want $A \otimes B$ to be graded anticommutative, we need a $-$ here, as $a \otimes b$, $c \otimes d \in (A\otimes B)_1$.

If we instead define \[ (a \otimes c)(b \otimes d) = (-1)^{\deg b\deg c} (ab \otimes cd) \] for homogeneous $b$ and $c$, the above example with $\deg a = \deg d = 0$, $\deg b = \deg c = 1$ gives \[ (a \otimes b)(c \otimes d) = (-1)^{1 \times 1}(ac \otimes bd) = -(ca \otimes db) = -(-1)^{0\times 0}(c \otimes d)(a \otimes b) = -(c\otimes d)(a \otimes b) \] as wished.

martini
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