15
$\begingroup$

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 $(a\otimes b)(c\otimes d)= (ac\otimes bd)$ ?

  • 0
    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..."2012-09-26
  • 0
    see Bourbaki, Algebras chapters 1 to 42012-09-26
  • 0
    Sorry, I had missed the graded bit.2012-09-26
  • 0
    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.2015-10-08

1 Answers 1