For A and B are simplicial abelian groups and NA and NB are the associated normalized chain complexesthen , there is a quasi-isomorphism $N(A\otimes B)\cong NA\otimes NB$NB\rightarrow N(A\otimes B)$, where the tensor product of simplicial groups is defined levelwise$(A\otimes B)_n=A_n\otimes B_n$and the tensor product of chain complexes is the usual$(C\otimes D)_n=\oplus_p (C_p\otimes D_{n-p})$. Thus, for example, a An associative simplicial ring leads directly in this way to a an associative differential graded ring. Another technical advantage of the normalized complex is its role in proving that homotopy groups of a simplicial abelian group are homology groups (Dold-Thom Theorem). 1 The question was about chains on a space, presumably singular chains. If what you're up to is, for example, learning or teaching the basic facts about singular homology, like homotopy-invariance and excision, then the switch to normalized chains seems like an unnecessary complication. In that context, you could say that non-normalized chains have the advantage. To expand on Ben's comments: If A and B are simplicial abelian groups and NA and NB are the associated normalized chain complexes then$N(A\otimes B)\cong NA\otimes NB$, where the tensor product of simplicial groups is defined levelwise$(A\otimes B)_n=A_n\otimes B_n$and the tensor product of chain complexes is the usual$(C\otimes D)_n=\oplus_p (C_p\otimes D_{n-p})\$. Thus, for example, a simplicial ring leads directly to a differential graded ring.