[Math] Tensor product of linear mappings versus chain complexes

Question

A chain complex of vector spaces $X_k$ is a sequence of linear mappings

$\dots \overset{d_{k-1}}{\longrightarrow} X_k \overset{d_{k}}{\longrightarrow} X_{k+1} \overset{d_{k+1}}{\longrightarrow} \dots$

such that $d_{k+1} \circ d_k = 0$. Here $k \in \mathbb N$ runs through the natural numbers. I have held the view that a chain complex $(X_k,d_k)$ is actually just a way to write down a pair of direct sums $(X,d) = ( \bigoplus X_k, \bigoplus d_k )$ that satisfy an additional property (namely, the graduation and the chain property).

However, the notions of tensor products that apply naturally to each of these points of view appear incompatible. Let $(X_k,d_k)$ and $(Y_l,D_l)$ be chain complexes. Then we define

$(X \otimes Y )_{p} := \bigoplus\limits_{k+l=p} X_k \otimes Y_l$

The tensor product of $d = \bigoplus d_k$ and $D = \bigoplus D_l$ in the sense of linear mappings is then defined by

$d \otimes D : X_k \otimes Y_l \rightarrow X_{k+1} \otimes Y_{l+1}$,

$( d \otimes D ) ( v \otimes w ) = ( d_k v \otimes D_l w )$

On the other hand, in the case of tensor products of chain complexes, we rather define

$d \otimes D : X_k \otimes Y_l \rightarrow X_{k+1} \otimes Y_{l} + X_{k} \otimes Y_{l+1}$,

$( d \otimes D ) ( v \otimes w ) = d_k v \otimes w + (-1)^k v \otimes D_l w$

The apparent incompatibility of these definitions contradicts my (likely simplified) view that chain complexes are only special notation for pairs of vector spaces and linear mappings. I expect this to become even more confusing as more structures are regarded (e.g., if the complex "is" a graded algebra).

Question: Is there an intuitive point of view, that there relates these two products? This is not discussed in literature (MacLane, Homology; Lang, Algebra).

Answer 1

Even at the risk of using slightly too heavy armory, I would like to shortly explain, how I would view the situation in a light, that naturally produces and "explains" the observations collected above:

• @Martin: Tthe natural extension on tensor products
• @TomGoodwillie: The "skew-Leibniz"-rule one has to use
• @JackHuizenga: the even-odd grading associated to it
• @Qiaochu: The mysterious "super-Lie-algebra"
• and in general the "additional" $d^2=0$ condition

First, being an algebraicist I "almost never" believe in maps $V\rightarrow V$. I rather always expect an entire algebra $H$ of operators acting on $V,W,...$ (e.g. universal enveloping of Lie algebras instead of Lie algebras or only single elements themselves). Furthermore as in this example, it sould be clearifies, how $H$ also acts on tensor products, the unit object $V=k$ and dual spaces .

Pretty much (and this is not entirely correct, but in a certain sense and with restrictions one can provo this, e.g. Etinghof's theorem) the above amount to $H$ is a Hopf algebra having additional structures:

• A coproduct $\Delta: H\rightarrow H\otimes H$ that tells you how to act on $V\otimes W$: Split up as $\Delta$ says, then act on each space $V,W$ as you already knew.
• A counit $\epsilon: H\rightarrow k$ telling you, with which scalar we act on the unit object $V=k$
• An antipode $S:H\rightarrow H$ that has to be used on an element $h\in H$ before letting it act in the argument of a linear form $f\in V^*$ (e.g. to get combined action again "the right way around")
• Such that... well pretty much of what you expect is true: You get again actions at all ($\Delta,\epsilon$ be algebra maps) and the so-defined action respects the re-bracking-iso of triple tensor products ("coassociativity"), the isomorphism $V\otimes k\cong V$ ("counitality") and the evaluation $V^*\otimes V\rightarrow k$ and dual basis ("antipode condition"). Respect means here, the action on the left side (e.g. in the second case via $(id\otimes \epsilon)\Delta$) matches the action on the right...or the isomorphism entertwines these two actions, or is a module homomorphisms or how you'd like to put it.

Take as examples (and if more interested read e.g. Susan Montgomory's book!):

• A group ring $k[G]$ with group elements $\Delta(g)=g\otimes g$, $\epsilon(g)=1$ and $S(g)=g^{-1}$, meaning just copy the element and act on each side of the tensor product as you wish (and inverse on "contragradiant" representations), as usual for group representation.
• A universal Lie-algebra enveloping $U(\ell)$ with Lie elements $\Delta(x)=1\otimes x+x\otimes 1$, $\epsilon(x)=0$ and $S(x)=-x$. As we expect: Lie algebras act on tensor products via Leibniz ;-)
• A quantum group such as the often discussed $U_q(\ell)$ or many exotic more discovered by Schneider/Andruskiewitsch ("Classifying pointed Hopf algebras with abelian coradical").
• Finally the Taft algebra for OUR purposes: $H_{Taft}=\langle g,x\rangle$ with $g^2=1$, $x^2=0$ and $gx=-xg$. Then $\Delta(g)=g\otimes g$ but now $\Delta(x)=g\otimes x+x\otimes 1$ a skew-LieElement/-derivation/-primitive.

It's the RadfordBiproduct/Majid/Bosonization of $k[\mathbb{Z}_2]$ with the BradiedHopfAlgebra/NicholsAlgebra $k[x]/(x^2)$, where the braiding is induced naturally by the sign-graduation and turns out to be the fermionic/Kozul-braiding $x^a\otimes x^b\rightarrow (-1)^{ab}x^b\otimes x^a$. For related more general cases see below... The Taft algebra has dimension only $4$ and somewhat the smallest nontrivial example - the noncommutativity of $g$ and $x$ has to exactly match skew-Leibniz-coproduct and the "premature" truncation $x^2=0$. Generally, Hopf algebras are pretty "picky" about their structures fitting together ;-) (ANSWER TO BELOW: if we take the subalgebra generated just by one such differential operator $x^2=0$, and require the graduation action to be faithful, and the Nichols algebra to be "indecomposable", we uniquely get the Taft algebra)

What does this have to do with the question?

Well, I find it natural to thing of a chain complex $(X_k,d_k)$ as a vector-space $V=\oplus_k X_k$ with an action of $H_{Taft}$ by $x=\oplus_k d_k$ acts differential ($x^2=0$!), $g$ as the odd/even operator $g.a=(-1)^{|a|}a$ (which you may want to refine to a $\mathbb{Z}$-graduation), both exactly anticommuting by the graduation shift.

And now the tensor product is right and "has-to-be" - satisfied with that?

$$x.(a\otimes b):=(x.a)\otimes (1.b)+(g.a)\otimes (x.b)=(x.a)\otimes b+(-1)^{|a|}a\otimes (x.b)$$

I would like to close with some remarks on the "super-Lie-algebras" and the general case:

The classification I mentioned roughly uses, that you may split a (pointed graded) Hopf algebra into (the "Radford-Biproduct") of a groupring (the ""coradical") and a "braided" Hopf algebra (i.e. in a category with a braiding induced by the group's conjugacy action and the groupelement-skewness). The generic case is that the braided Hopf algebra is like the universal enveloping of a Lie algebra, but in these braided "Nichols algebras" we may have truncations. When and how is pretty tricky and open especially over nonabelian groups, while over abelian groups, they're nowadays well understood by the works of Heckenberger. You can image them (and the theory of root systems carries overs!!) as Lie algebras in a braided sense, and this allows specific but sometimes much more exotic Dynkin diagrams - in the easiest case of group $\mathbb{Z_2},\mathbb{Z_3}$ these are e.g. in physics called Super- and Color-Lie-algebras, but of course there are much more interesting cases ;-) ;-)

I you have to have more, you might want to check out my article "Nichols algebras" on Wikipedia ;-) ;-) and run over "our" Nichols algebra $k[x]/(x^2)$ and braiding underlaying the Taft algebra.