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The tensor product of chain complexes (of $R$ modules) $C_\bullet ,D_\bullet$ is defined as $$(C_\bullet \otimes D_\bullet )_n = \bigoplus_{i+j=n} C_i \otimes_R D_{j}$$

I understand this definition works by yielding a chain complex (with a nilpotent boundary operator as usually defined), but I don't understand anything about it beyond that.

Questions:

  1. What is the motivation behind this definition?
  2. Is there any geometric insight to be gained here?
  3. Where can I read about its history?

Added: Does the tensor product of chain complexes as defined above posses some universal property (which is easy to formulate) in $\mathsf {Ch}_\bullet$?

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  • $\begingroup$ check Jean Dieudonné's "A History of Algebraic and Differential Topology 1900-1960". $\endgroup$ – janmarqz Dec 22 '14 at 17:36
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    $\begingroup$ @janmarqz I looked (around page 92), but the definition is just given there :( $\endgroup$ – user153312 Dec 22 '14 at 17:45
  • $\begingroup$ If you're computing (co)homology of topological spaces, tensor product of complexes corresponds to the usual product of spaces. $\endgroup$ – Grigory M Dec 22 '14 at 17:47
  • $\begingroup$ @Exterior: did you see around that it is used for the Künneth formula? $\endgroup$ – janmarqz Dec 22 '14 at 17:47
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    $\begingroup$ @Exterior Details depend on the version of (co)homology you're using — but if, say, $X$ and $Y$ are CW-spaces then $C(X\times Y)\cong C(X)\otimes C(Y)$ where $C(-)$ is the cellular chain complex. $\endgroup$ – Grigory M Dec 22 '14 at 17:57
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Let $(C,d_C)$ and $(D, d_D)$ be two chain complexes of $R$-modules, where $d_C$ and $d_D$ are differentials of degree $+1$. By definition, each $C_i$ and $D_j$ are $R$-modules. We want to "compose" the above complexes in a tensorial way; the definition you propose has 2 main effects:

  1. each $(C\otimes_R D)_n$ is again an $R$-module.
  2. the induced differential (which is probably the missing part in the OP) $$d_{C\otimes_R D}:=d_C\otimes_R 1_D + 1_C\otimes_R d_D,$$

is compatible with the (co)homological grading. In fact, for all $c\in C_i$ and $g\in D_j$, s.t. $i+j = n$, i.e. $c\otimes_R g\in (C\otimes_R D)_{n}$ then

$$d_{C\otimes_R D}(c\otimes_R g)= d_C c\otimes_R g + (-1)^i c \otimes_R d_Dg\in (C\otimes_R D)_{n+1}, $$

as $d_C c\in C_{i+1}$ and $d_D g\in D_{j+1}$. We used the Koszul sign rule.

For topological / geometric insights I refer to the text "Rational Homotopy Theory" by Felix, Halperin and Thomas. For formal definitions and applications in homological algebra the book by Gelfand and Manin is recommended, instead.

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  • $\begingroup$ I don't know what the Koszul sign rule is, and a quick google search yielded only this extremely brief sketch of the idea. $\endgroup$ – user153312 Dec 22 '14 at 18:11
  • $\begingroup$ The Koszul sign rule justifies the sign $(-1)^i$ in the formula of the differential for the tensor complex above; it is not so relevant at this stage. The link you refer to explains the rule quite well: it can be summarized in ""exchanging two symbols you multiply the result by the sign (−1), raised to the power of the products of their degrees". $\endgroup$ – Avitus Dec 22 '14 at 18:26
  • $\begingroup$ It's frustrating to just accept that though. Can you direct me to more detailed explanations? $\endgroup$ – user153312 Dec 22 '14 at 18:28
  • $\begingroup$ Uhm, ok: try to compute $d_{C\otimes_R D}^2(c\otimes_R d)$ without extra $(-1)^{...}$ sign in front of $c\otimes d_D d$: do you get $d_{C\otimes_R D}^2(c\otimes d)= 0$, as wished? Answer: no. This implies that $C\otimes_R D$ is not a chain complex. The Koszul sign rule guarantees $d_{C\otimes_R D}^2(c\otimes d)= 0$, instead. $\endgroup$ – Avitus Dec 22 '14 at 18:32
  • $\begingroup$ Using $d$ for differential and for element of $D$ is very confusing. Could you alter your notation in your answer? $\endgroup$ – tom Dec 22 '14 at 19:04
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$\{Ci\otimes Dj\}$ is nice double complex,with left hand side is $n$-th object in total complex of the double complex. you can refer to balancing ext and tor in Weibel's book. it has a nice application to prove two left derived functor of $A\otimes B$ gives same functor.

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Ok I have never seen definition of tensor product of chain complexes before but it seams to me quite sensible. I did a very crude drawing which might give you some motivation behind the definition (If I got it right :D)

In the picture we have two chain complexes $C_\bullet,D_\bullet$, in fact they are simplicial complexes of dimension $1$. They are both just an edge with two end vertices. Now you would like to somehow combine them and get square and that is the tensor product. Or you could take two circles and get torus or circle and edge and get cylinder.

I did not label every element of $C_\bullet \otimes D_\bullet$ I just labeled vertex $a\otimes d$, edge $e\otimes c$ and face $e\otimes f$.

enter image description here

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    $\begingroup$ I would love an explanation of the image :) $\endgroup$ – user153312 Dec 22 '14 at 19:30
  • $\begingroup$ Is that enough? $\endgroup$ – tom Dec 22 '14 at 19:44
  • $\begingroup$ I don't follow. Why are chain complexes simplicial complexes? What does the definition I gave have to do with this? $\endgroup$ – user153312 Dec 22 '14 at 20:13
  • $\begingroup$ This is just simple example of tensor product between two very simple simplicial complexes. $\endgroup$ – tom Dec 22 '14 at 20:18
  • $\begingroup$ Okay, but I don't understand what this square has to do with tensor products in particular.. It could just as easily be a representation of a homotopy, no? Why does your illustration match the definition above? $\endgroup$ – user153312 Dec 22 '14 at 20:25

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