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Stream: theory: algebraic topology

Topic: external vs internal homology

Tim Hosgood (May 17 2021 at 18:20):

There is a really nice way of thinking about homology of spaces (as explained somewhere on the nLab) as just "$\infty$ mapping spaces" between "spaces", and then we can recover the usual notion via some classical model categorical stuff: $H(X,A)$ can be computed by taking a (co)fibrant resolution (e.g. the Čech nerve of $X$ or some projective resolution of the sheaf/space/group $A$). I like this story a lot, and it gives me a nice way to think about things.

But then we have the idea of internal homology (as opposed to the external notion described above). This is where we look at a chain complex and construct a new chain complex (which is chain homotopic to the original) by looking at cycles modulo boundaries. We can use this internal homology to calculate external homology sometimes (e.g. external homology of a space $X$ with coefficients in $A$ given by the internal homology of the singular chain complex of $X$ with coefficients in $A$), but I don't really know a story about this internal homology analogous to the one I described above for external homology. Maybe there is something nice to say that goes through the notion of formal complexes (i.e. those which are quasi-isomorphic to their internal homology)? Maybe this whole story is somehow analogous to the story of internal vs. external homs? (that would be really satisfying).

I know this is rather a vague question, but does anybody have any insight to offer? :smile:

Tim Hosgood (May 17 2021 at 18:22):

this whole thought was prompted by the fact that we can write "chain homotopic maps induce identical maps in homology" as "$H(Hom(C_\bullet,D_\bullet))\subseteq Hom(H(C_\bullet),H(D_\bullet))$", where the inclusion is an equality under certain very nice conditions (e.g. every module is projective, or we work over a field)

Reid Barton (May 17 2021 at 18:46):

If you start with a chain complex of abelian groups then its homology groups are also the homotopy groups of the corresponding simplicial abelian group (seen as a simplicial set). Is that the sort of thing you're looking for?

Fawzi Hreiki (May 17 2021 at 18:57):

This is the Dold-Kan correspondence right?

Fawzi Hreiki (May 17 2021 at 18:58):

Yuri Manin's book on homological algebra has a very simplicial homotopic flavour

Fawzi Hreiki (May 17 2021 at 18:59):

'Methods of Homological Algebra' I beleive

Mike Shulman (May 17 2021 at 18:59):

In a closed monoidal $\infty$-category $C$ you can build the "internal cohomology" $H^n(X,A)$ of an object $X$ with coefficients in another object $A$ by taking the internal-hom $[X,\Omega^{-n}(A)]$ and 0-truncating it (assuming $C$ has 0-truncations and $A$ has a $(-n)$-fold delooping, the latter being automatic if $n\le 0$ or if $C$ is stable). This gives a sequence of objects indexed by the external $\mathbb{N}$, however. In a stable world you could do something like suspend each $H^n(X,A)$ $n$ times to put it in the right dimension and then take the product of all of them, to get a single object that includes all the data.

Fawzi Hreiki (May 17 2021 at 19:00):

Mike Shulman said:

In a closed monoidal $\infty$-category $C$ you can build the "internal cohomology" $H^n(X,A)$ of an object $X$ with coefficients in another object $A$ by taking the internal-hom $[X,\Omega^{-n}(A)]$ and 0-truncating it (assuming $C$ has 0-truncations and $A$ has a $(-n)$-fold delooping, the latter being automatic if $n\le 0$ or if $C$ is stable). This gives a sequence of objects indexed by the external $\mathbb{N}$, however. In a stable world you could do something like suspend each $H^n(X,A)$ $n$ times to put it in the right dimension and then take the product of all of them, to get a single object that includes all the data.

Where could someone go to read more about this?

Tim Hosgood (May 17 2021 at 19:02):

Fawzi Hreiki said:

This is the Dold-Kan correspondence right?

oh yes, I guessed that Dold–Kan might pop up somewhere in this story

Tim Hosgood (May 17 2021 at 19:03):

Mike Shulman said:

In a closed monoidal $\infty$-category $C$ you can build the "internal cohomology" $H^n(X,A)$ of an object $X$ with coefficients in another object $A$ by taking the internal-hom $[X,\Omega^{-n}(A)]$ and 0-truncating it (assuming $C$ has 0-truncations and $A$ has a $(-n)$-fold delooping, the latter being automatic if $n\le 0$ or if $C$ is stable). This gives a sequence of objects indexed by the external $\mathbb{N}$, however. In a stable world you could do something like suspend each $H^n(X,A)$ $n$ times to put it in the right dimension and then take the product of all of them, to get a single object that includes all the data.

this idea of "internal" is different from the one that I describe though, right? my main confusion in the analogy is how internal cohomology has no choice of coefficients anywhere

Mike Shulman (May 17 2021 at 20:32):

Oh, well, for something more like your "internal homology" you would do the same thing with the plain 0-truncations of $X$. A general notion of "internal homology with coefficients" would be to first tensor with the coefficients object $X\otimes A$ and then truncate; your homology has "coefficients in the unit object".

Tim Hosgood (May 17 2021 at 20:40):

hmm, I don't understand I'm afraid. are you saying that we can recover the homology of a chain complex $X$ by looking at $[X,\Omega^{-n}(I)]$ somehow (where $I$ is the unit object for chain complexes)?

Mike Shulman (May 17 2021 at 20:44):

No, $X\otimes \Omega^{-n}(I)$. The internal-hom is for cohomology.

Tim Hosgood (May 17 2021 at 20:49):

so, just to reiterate, given a chain complex $C_\bullet$ of $R$-modules, we recover the $H_n(C_\bullet)=\operatorname{Ker}d_n/\operatorname{Im}d_{n+1}$ by taking $C_\bullet\otimes\Omega^{-n}(R[0])$?

Mike Shulman (May 17 2021 at 20:53):

And then the 0-truncation. (I'm talking about bounded-below complexes here.) I think this is really tautological since the 0-truncation is just $H_0$, and that tensor just shifts things down so that $H_n$ becomes $H_0$.

Tim Hosgood (May 17 2021 at 20:57):

ah, of course!

Tim Hosgood (May 17 2021 at 20:58):

so it really is the case that external vs internal homology is somehow analogous to external vs internal homs. lovely!

Tim Hosgood (May 17 2021 at 20:58):

is there a nice reference for this point of view in general? (besides HTT I suppose...) as in, somewhere that might write down the definition of the homology of a chain complex as the abstract definition

Mike Shulman (May 17 2021 at 20:59):

Hmm, not that I know of really.

Tim Hosgood (May 17 2021 at 21:03):

I guess I'd like to relearn homological algebra, but from the model category/$(\infty)$-categorical POV

Tim Hosgood (May 17 2021 at 21:06):

is there a nice reason (from this point of view) as to why "$H(Hom(C_\bullet,D_\bullet))\subseteq Hom(H(C_\bullet),H(D_\bullet))$", where the inclusion is an equality under certain very nice conditions (e.g. every module is projective, or we work over a field)?

Mike Shulman (May 17 2021 at 21:17):

I don't know of one.

Tim Hosgood (May 17 2021 at 21:47):

ok, thanks for all your other insight anyway though!