A Mind for Madness

Musings on art, philosophy, mathematics, and physics


1 Comment

Spectral Sequence Arrows Revisited

I don’t want to do much today. In fact, we’ll probably not cover any new ground, but start to fill in what is going on a little better. Notice that in order to work with the spectral sequence last time, we didn’t actually have to figure out what the {i, j,} or {k} or {d} maps were. We only needed that they existed. The main thing of today is to work out all these sequences. Some are exact. Some are chain complexes. Some are commutative diagrams. Most of them fit together in a really organized way, other parts are just related. It is useful to see how these fit together, since sometimes you can get by without the maps as long as you know it is part of a commutative exact diagram.

Recall that our current situation is that given a filtered chain complex {\cdots K^{p-1}\subset K^p\subset \cdots \subset K} we get an exact couple {(D_{p,q}, E_{p,q})} where {D_{p,q}= H_{p+q}(K^p)} and {E_{p,q}=H_{p+q}(K^p/K^{p-1})} by taking the long exact sequence in homology associated to the short exact sequence {0\rightarrow K^{p-1}\rightarrow K^p\rightarrow K^p/K^{p-1}\rightarrow 0}.

Two times ago we saw exactly what the {i_{p,q}}, {j_{p,q}} and {k_{p,q}} maps were.

Let’s develop them again, but being more careful as to how they all fit together. Well {i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})} and it comes from the induced maps on homology from the injection {K^p\hookrightarrow K^{p+1}}. So we get infinite strings of {i_{p,q}} for fixed {q} that relate as injections {p} changes. i.e we’ll get as part of a large diagram columns that look like {\cdots \rightarrow H_{p+q}(K^{p-2})\rightarrow H_{p+q}(K^{p-1})\rightarrow H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})\rightarrow\cdots}.

The {j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})} are induced on homology from the projection map {K^p\rightarrow K^p/K^{p-1}}.

The {k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})} comes from the snake lemma. It is the boundary map in the long exact sequence.

Note that these last two will fit into rows: {\cdots \rightarrow H_{p+q}(K^p)\stackrel{j}{\rightarrow} H_{p+q}(K^p/K^{p-1})\stackrel{k}{\rightarrow} H_{p+q-1}(K^{p-1})\stackrel{j}{\rightarrow} H_{p+q-1}(K^{p-1}/K^{p-2})\rightarrow\cdots}.

Thus we can form a large commutative diagram from these long exact sequences, and figure out how to extrapolate the exact couple from it. Below is the diagram relating what was just said:

(Sorry about the image. I’ll figure this out eventually. For now you can click on it and zoom to get the full size).

The labelled arrows are the {i,j,k} maps of just a single exact couple. Note that there are infinitely many exact couples going on here, and they are all related. Note also that the exact couples don’t just go across or down.

Now let’s figure out the differentials and how to get the spectral sequence. Well, take the {H_{p+q}(K^p/K^{p-1})} spot. Recall that in the exact couple, this is the {E} term which is where our chain complex comes from. Then doing {j\circ k} is the {d} map. Note this is the really the {d^1} map. So our page of {E^1_{p,q}} comes from this row only.

Take homology with respect to this, and due to the nature of the exact couples, we’ll get another page of these things. Now we are on the {E^2_{p,q}} page of groups. The {d^2} map now is to go right, then up, then right. This is the composition {k\circ i^{-1} \circ j: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-2}/K^{p-3})}.

Take homology again. In general the {d^r} map is going right, then up {r-1} times then right one more time. This gives us exactly what we said last time, {d^r_{p,q}: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}.

Hopefully this helps clarify some of the inner workings of what is going on here. Next time we’ll talk about why the conditions from last time imply that the spectral sequence converges. We won’t prove it, but there are some nice concepts behind why it works that should be enlightening.


Leave a comment

Serre Spectral Sequence

Today we’re going to back up and not worry about the tediousness of the last post. Let’s try to form some motivation and see how these things work even if we have to just assume some of the theory works for now. I think it will be helpful. So why care about spectral sequences? Well, one thing is that sometimes it is really hard to compute say {H_n(C)}, but it isn’t so bad to do {H_n(C\otimes A)} for some {A}‘s. There is no good way to convert the information you get out of {H_n(C\otimes A)} to information about {H_n(C)}, but this is exactly what the Bockstein spectral sequence does.

Today, I’ll talk about the Serre spectral sequence. This will help us calculate the homology of the loop space of an {n}-sphere. There is a nice proof that Tor is independent of which resolution you take (i.e. {Tor(A, B)} can be calculated by resolving {A} or {B}). There is a nice proof of the Snake Lemma using spectral sequences. There is a proof of the K\:{u}nneth formula. For the algebraic geometers, the Leray spectral sequence tells us information about sheaf cohomology. Anyway, there are tons of examples hitting many areas of math of uses of spectral sequences.

Right now we’ll not worry about the technical details of when and how a spectral sequence converges, but recall that the situation we ended on was that given a complex {K}, we could form a spectral sequence from a filtration {\cdots \subset K^p\subset K^{p+1}\subset \cdots \subset K}. We’ll just accept for now that if we have {K^p=0} for {p\max(p, q+1)} then {E^r_{p,q}\simeq E^\infty_{p,q}}. This {E^\infty} term is the associated graded group of a filtration of {H_*(K)}.

This should be exciting. It means that if we can formulate a statement that reads, “There is a spectral sequence with {E^2_{p,q}=} (fill in the blank), differential map {d^r: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}, that converges strongly to {H_{p+q}}(fill in again).” Then we can just chase around differentials and hope that everything collapses fast (lots of times these things only have a few terms or even {E^2=E^\infty}). If this happens then we can just read off what the known things are, and we’ll have figured out new information. Note that being able to say such a statement is not bad. We only need the mere existence to begin work, and taking those conditions above for granted, we actually can formulate lots of statements such as these.

Let {\Omega S^n} be the loop space of {S^n}, namely the space of loops based at the north pole. Let {PS^n} be the path space, the space of paths starting at the north pole. Note here that our standard tools of algebraic topology are not very useful in trying to calculate {H_q(\Omega S^n)}. But we know that {PS^n} is contractible using the obvious map of retracting along all the paths simultaneously. And we also know something very useful, that there is a spectral sequence with {E^2_{pq}=H_p(S^n)\otimes_\mathbb{Z} H_q(\Omega S^n)}, differentials {d^r : E^r_{pq}\rightarrow E^r_{p-r, q+r-1}} converging strongly to {H_{p+q}(PS^n)}.

So recall that we get an entire page of groups. The {E^2} groups ({r=2}). Since {H_p(S^n)} is {0} for all {p} except {p=0, n}, and is {\mathbb{Z}} in those spots. We also know that {H_0(\Omega S^n)=\mathbb{Z}}. This allows us to fill in the whole {q=0} row, and all {p} columns are completely {0} except in {p=0} and {p=n}.

The differential map {d^r} goes to the left {r} and up {r-1}.

Examine the bottom row. Due to all the {0}‘s in columns not {p=0} or {n}, all the differential maps that don’t land in the {p=0} column must be 0. Thus there is only one possible non-zero differential coming out of there. Namely, the {d^n_{n,0}} map. It lands in the {E^2_{0, n-1}=H_{n-1}(\Omega S^n)} spot. Now the complex of {d^n} maps is {\cdots \rightarrow 0\rightarrow \mathbb{Z}\stackrel{d^n_{n,0}}{\rightarrow} H_{n-1}(\Omega S^n)\rightarrow 0\rightarrow \cdots }. Recall that to get to the different {r} values in {E^r_{p,q}} you take homology with respect to the {d^r} maps. Since {E^\infty_{p,q}=H_{p+q}(PS^n)=0} for {p+q\neq 0} we must have {d^n_{n, 0}} an isomorphism in order for it to vanish. This gives us a {\mathbb{Z}} in the 0, {n-1} spot and hence another in the {n}, {n-1} spot. We can keep repeating this argument giving a {\mathbb{Z}} in the {p=0, n} and {q=} multiples of {n-1} spots.

Since the differentials are all {0} before {r=n}, nothing changes for {E^2=E^3=\cdots E^n}. Then at {r=n} we get isomorphisms except at the {E^2_{0,0}} position, so everything vanishes except that position which stays a {\mathbb{Z}}. Thus {H_q(\Omega S^n)=\begin{cases} \mathbb{Z} \ if \ (n-1)|q \\ 0 \ else\end{cases}}.

This is a pretty typical situation. Hopefully that gives a better overview of the process so that going back to the general case will have some sort of reference. If there are questions, feel free to ask. Maybe I’ll do another example before going back to theory.


3 Comments

Exact Couples

Today we begin a long set of posts on spectral sequences. I must first off say that I only learned what a spectral sequence was about 2 weeks ago, so I am far from knowledgable about them. This means if you have questions, I may not know the answers. I’m also going to go incredibly slowly. I want to carefully develop everything and give lots of motivating examples. Lastly, all of this is coming from the notes I’ve taken from John Palmieri’s class this quarter, so any great insights should be attributed to him and mistakes should be attributed to myself.

Like I’ve already said, I’m not familiar with the literature, so I’m not sure what the standard development is. For people who know some of this already, we’re going to define the spectral sequence associated to an exact couple. Then anytime we want a spectral sequence, we just need to figure out the exact couple it comes from. This is a different approach than say Weibel.

In general, this is going to work in any abelian category, but we’ll use the category of {A}-modules just to keep it less scary for those that haven’t worked with abelian categories.

Our first definition is that of an exact couple. This is a diagram of {A}-modules, {D} and {E} such that:

is exact in every spot. Then we immediately get a map {d: E\rightarrow E} by {d=j\circ k}. This may look strange, since we’re not just going around the diagram. We sort of jump, but this is fine since at this stage we aren’t identifying {i(D)\subset D} or anything. Now this map has a nice property, {d\circ d=j\circ k\circ j\circ k=j\circ (k\circ j)\circ k=j\circ (0)\circ k=0} by exactness that middle term is {0} making the whole thing {0}.

This means that {\cdots \stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\rightarrow \cdots} is actually a chain complex. Thus we can take homology: \displaystyle {H(E, d)=\frac{ker d}{im d}}.

Now take {D'= im i}. Define the map {i': D'\rightarrow D'} by {i'=i\big|_{D'}}. Take {E'=H(E, d)}. We will now think of {j': D'\rightarrow E'} as {j\circ i^{'-1}}, but to be precise we pick an element whose preimage is {x=i^{-1}(y)}, and take an equivalence class, so {j'(x)=[j(y)]}. Lastly define {k': E'\rightarrow D'} by {k'([z])=k(z)}. Now we have made lots of choices, but as usual in homological things it turns out by examination that everything is well-defined. It is also the case that with these new maps, the corresponding diagram is exact at every place. We call this the derived exact couple.

Since the derived exact couple is an exact couple, there is nothing stopping us from defining the derived exact couple of the derived exact couple. Then do it again, and again, … . Obviously, if we aren’t careful we’ll lose track of where we are, so choosing notation is essential. Suppose {(D, E)} form an exact couple. Then let {D^1=D} and {E^1=E}. Next let {D^2=D'} and {E^2=E'}. This inductively gives us that {D^{r+1}=(D^r)'} and {E^{r+1}=(E^r)'}. In each of these we get three maps which tell us that each stage is an exact couple {i^r: D^r\rightarrow D^r}, also {j^r: D^r\rightarrow E^r}, and {k^r: E^r\rightarrow D^r}.

Since the triangle is exact, defining (which technically already happened in the definition of {E^{r+1}}) {d^r: E^r\rightarrow E^r} by {d^r=j^r\circ k^r} we get a chain complex and can take homology. We say that {\{(E^r, d^r)\}_{\{r\geq 1\}}} is the spectral sequence associated to the exact couple. Note that {E^{r+1}=H(E^r, d^r)}.

This is probably incredibly useless for people wanting to know what a spectral sequence is and how it is used. But in general we are going to have a lot more information floating around than just this abstract exact couple. More specifically, we have to get the exact couple somehow, and the process of forming it will give us things to work with. The other thing is that we want to know that {\displaystyle \lim_{r\rightarrow\infty}(E^r, d^r)} converges in some sense to something, and hopefully we have information about what this “{E^\infty}” term is.

I don’t want to go through an entire example today, but we’ll set up the situation for the example (for those who have seen this, it will be the Serre Spectral Sequence).

Suppose we have a chain complex {K}. Then given a filtration {\cdots \subset K^n\subset K^{n+1}\subset \cdots \subset K} (not necessarily starting at the 0 complex or ending at the total complex, but the {j}th grade of any {K^n} should be a submodule of the {j}th grade of {K} and the submodules should respect the differentials). Then we will get a short exact sequence of chain complexes {0\rightarrow K^{P-1}\rightarrow K^P\rightarrow K^P/K^{P-1}\rightarrow 0}. This is just because it works trivially on each grade.

Well, our old standby says that a short exact sequence of chain complexes induces a long exact sequence in homology, so {\cdots \rightarrow H_n(K^{P-1})\stackrel{i}{\rightarrow} H_n(K^P)\stackrel{j}{\rightarrow} H_n(K^P/K^{P-1})\stackrel{k}{\rightarrow} H_{n-1}(K^{P-1})\rightarrow \cdots}.

From here we define our exact couple. Let {D_{p,q}=H_{p+q}(K^p)} and {E_{p,q}=H_{p+q}(K^p/K^{p-1})}.

Now use the maps given in the long exact sequence.

So {i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})} which is really {D_{p,q}\rightarrow D_{p+1, q-1}}.

Our {j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})} which maps {D_{p,q}\rightarrow E_{p,q}}.

Lastly the map {k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})} which maps {E_{p,q}\rightarrow D_{p-1, q}}.

Lastly, the differential {d_{p,q} = j_{p-1, q}\circ k_{p,q}: E_{p,q}\rightarrow E_{p-1, q}}.

Don’t worry if this seems horribly confusing. We’ll look at a very concrete example next time in which we’ll see that often everyting stays stationary or goes to 0 leaving us with very little. I’ll just try to summarize a little of this filtered chain complex example. We are getting infinitely many exact couples all mingled together somehow. If we think of a {(p,q)}-plane, then at each integer valued place we get a group {E_{p,q}}. The differential maps move us to the “left” on place. So we have rows of chain complexes for each fixed {q}. Remember that when we form the spectral sequence this is really only the {E^1_{p,q}} part. We can think of this term as an entire “sheet” or “page” of groups related to each other. We get a page {E^r_{p,q}} for each {r\geq 1}.

Some things that would be nice (and to get you thinking about why this might be useful) is if for each fixed pair {(p,q)} the groups stabilize, i.e. {E^N_{p,q}=E^{N+1}_{p,q}=\cdots}. Then we would have a notion of {E^\infty_{p,q}}. It would be nice to know how this limit term relates to the original chain complex, or filtration. It would be nice if the spectral sequence limit were somehow independent of the filtration, that way taking a limit of different filtrations would give us information we might not know. Like I said earlier, don’t worry if this post was confusing. It just needed to be done. After some examples, this will probably not seem so bad.


6 Comments

Abelianization of the Fundamental Group

I guess I have no reason to offer explanation for lack of posting, but in general this has been one of the best weeks ever and at the same time one of the worst. The worst because I’ve been fairly ill and can’t seem to fully conquer it. It has been the best week for reasons I won’t mention, since I try to keep personal stuff out of this blog as much as possible (but if you know of my other blog which is purely my personal stuff, then you can read about it to your heart’s content, but I refuse to give any hints at all as to how to find that). Both of these factors has lead to a fairly unproductive week.

I may take a more algebraic topology approach for awhile. This is mainly since I’m doing a reading course on Hatcher (with two other students), and before I go present stuff to them and the prof I want to clarify my ideas.

Tomorrow I’m presenting the proof that H_1(X)\cong \pi_1(X)^{ab} for path connected spaces. This is a pretty wonderful result if you think about it. We have exactly how first homology and the fundamental group relate. In fact, the first thing you’d think to do (granted, this might take a little while) is the thing that works.

We can naturally think of paths and singular 1-simplices as the same thing, since they are both just continuous maps to the space out of a closed interval. So after rescaling, a loop f:[0,1]\to X is actually also a 1-cycle since \partial f=f(1)-f(0)=0.

The overall idea of this proof is then to show that h: \pi_1(X, x_0)\to H_1(X) is a well-defined homomorphism with image all of H_1(X) and kernel the commutator subgroup. Almost all of these facts are fairly straightforward.

First, we’ll need a few ways in which our different modes of thinking about loops versus 1-cycles correlate. If as a path f\equiv c, a constant, then f\sim 0 the cycle is homologous to 0. If two paths are homotopic (in the path homotopic and hence equivalence class of \pi_1(X) sense), denoted f\simeq g then they are homologous. Concatenation of paths (and hence the operation in the fundamental group) is homologous to addition of the cycles (the operation in first homology). Lastly, traversing a path backwards is homologous to negating the cycle: \overline{f}\sim -f.

So we’ll use these four facts without proof, since they are fairly standard and the proof is long enough as it is.

Recall the definition h([f])=f. The second fact, gives us that h is well-defined since any other representative of the equivalence class will be homotopic to the original, and hence the outputs will be homologous.

The third fact gives us that h is a homomorphism of groups.

Our first bit of effort comes from showing that h is surjective. Here we will use the path-connected hypothesis (everything else so far is true without it). Let \sum n_i\sigma_i be any 1-cycle. We must construct a loop that maps to it.

Since the n_i are integers, we can assume each is \pm 1 by just repeating the \sigma_i as many times as needed. But all the \sigma_i with -1 in front can be replaced by -\overline{\sigma_i} by the fourth property. This converts all the n_i to 1. Thus \sum n_i\sigma_i\sim \sum \sigma_k.

But \partial(\sum \sigma_k)=0, so all the endpoints must cancel. So for any \sigma_k that is not a loop, in order to cancel \sigma_k(1), there must be a \sigma_j such that \sigma_j(0)=\sigma_k(1). i.e. there is some \sigma_j that we can concatenate with to form \sigma_k\cdot \sigma_j. In order to cancel the \sigma_k(0) some other \sigma_j must exists with endpoint \sigma_j(1)=\sigma_k(0).

So we can concatenate, then rescale, and group all of these cycles into a collection of loops by the third property. So the only remaining thing we must do is get it to be a single loop. But X is path-connected, so pick some basepoint x_0\in X. For any of these possibly disjoint loops floating around, we can pick a basepoint at each and connect with a path \gamma_i from x_0 to the basepoint of the i-th loop. By the third and fourth properties \gamma_i\cdot \sigma_i\cdot \overline{\gamma_i}\sim \sigma_i. So now all loops start and end at x_0 and we can combine into a single loop \sigma. Thus h([\sigma])=\sum n_i\sigma_i.

Now comes the hard part. We want ker h=\pi_1(X, x_0)'. The one containment is easy. Since H_1(X) is abelian, by the universal property of the commutator subgroup, \pi_1(X)'\subset ker h. The method to get the other direction is to show that for any h([f])=0, we must have that [f] is trivial in the abelianization.

Suppose [f]\in\pi_1(X) such that h([f])=0. Since f is a cycle, there is some 2-chain \sum n_i\sigma_i such that \partial (\sum n_i\sigma_i)=f. So as before, we can assume each n_i=\pm 1. Now the goal is to associate a 2-dimensional \Delta-complex to \sum n_i\sigma_i by taking for each \sigma_i a \Delta_i^2 and identifying pairs of edges which we’ll call K.

So before writing this process down, we should examine what the process will be geometrically. It turns out that K will be an orientable compact surface with boundary, since we are just fitting together a finite collection of disjoint 2-simplices (this is not meant to be obvious). The component containing the boundary is a closed orientable surface with an open disk removed. Since connected sums of tori can be expressed as a 2n-gon with pairs of edges identified in the manner aba^{-1}b^{-1}, cdc^{-1}d^{-1} etc, we see that f is homotopic to a product of commutators.

Writing this in detail algebraically is much trickier. Given any \sigma_i, we have \partial \sigma_i=\tau_{i0}-\tau_{i1}+\tau_{i2}, where \tau_{ij} are singular 1-simplices. Thus f=\partial(\sum n_i\sigma_i)=\sum (-1)^j n_i\tau_{ij}.

Keep the picture of a triangle in your head. When we fit together the triangles we are getting pairs of edges. The signs on these pairs are opposite and so will cancel when we sum. The remaining (of the three sides) \tau_{ij} is a copy of f. This forms our \Delta-complex K.

Now form \sigma : K\to X by fitting together the \sigma_i maps. Deform \sigma relative the edges that correspond to f by mapping each vertex to x_0. So we have a homotopy on the union of the 0-skeleton with edge f, so by the homotopy extension property we get a homotopy on all of K.

Now restrict \sigma to the simplices \Delta_i^2 to get a new chain \sum n_i\sigma_i with boundary f and \tau_{ij} loops at x_0.

Now we just need to check whether the class is trivial or not: [f]=\sum (-1)n_i[\tau_{ij}]=\sum n_i [\partial \sigma_i] where [\partial \sigma_i]=[\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]. But \sigma_i gives a nullhomotopy of \tau_{i0}-\tau_{i1}+\tau_{i2} and we are done.

Thus ker h=\pi_1(X, x_0)' and by the First Iso Theorem we have H_1(X)\cong \pi_1(X, x_0)^{ab}.


1 Comment

Derived Functors II

First off, we need to get this pesky result out of the way that derived functors are independent of the choice of resolution. So we do this by proving a related result. Suppose \cdots\rightarrow F_i \stackrel{\phi_i}{\rightarrow}F_{i-1}\rightarrow \cdots \rightarrow F_1\stackrel{\phi_1}{\rightarrow} F_0 is a a complex of projective A-modules, and \cdots G_1\stackrel{\psi_1}{\rightarrow} G_0 is a complex of A-modules. Let M=coker \ \phi_1 and N=coker \ \psi_1. Suppose the homology of G vanishes except H_0(G)=N. Then every map \beta\in Hom_A(M, N) is a map induced on H_0 by a map of complexes \alpha : F\to G and is determined up to homotopy by \beta.

Before proving this, note that as a corollary we get that any two projective resolutions are homotopy equivalent, and hence the derived functors have constructed on different resolutions have a natural isomorphism between them.

Proof: I knew I should never have tried to do homological algebra without a good way to do diagrams on wordpress. This is clearer if you draw it out…but the idea for existence is to inductively lift your maps. Lift F_0\to M\to N to \alpha_0: F_0\to G_0, then \alpha_0\phi_1: F_1\to ker(G_0\to N)=im(G_1\to G_0). Thus we lift this to \alpha_1: F_1\to G_1 and continue this process. This gives the map of the complexes.

We now want uniqueness up to homotopy. Suppose we have two lifts of \beta: M\to N say \alpha and \gamma. Then \alpha - \gamma lifts the zero map. i.e. we need only show that any lifting of 0 is homotopic to 0. Suppose then that \eta is a lifting of zero. We need that \eta_i=h_{i-1}\phi_i + \psi_{i+1}h_i for some h_i: F_i\to G_{i+1}. Note that \eta_0 : coker \ \phi_1\to coker \ \psi_1 takes F_0\to im\psi_1. So we lift to h_0: F_0\to G_1 such that \psi_1 h_0=\eta_0. But now \psi_1(h_0\phi_1-\eta_1)=\eta_0\phi_1-\psi_1\eta_1=0. Thus h_0\phi_1-\eta_1 maps into ker \ \psi_1=im \ \psi_2. But F_1 is projective so we can lift to h_1: F_1\to G_2. Repeat this process.

I highly recommend just doing the diagram chasing yourself. This is sort of a mess to read, and so should only be used as a sort of guideline if you get stuck somewhere.

Hmm…what else did I say I would do? Oh right. If you’ve seen homological constructions, then you can probably guess that there is a connecting homomorphism type theorem. i.e. Something that is phrased, “a short exact sequence of complexes induces a long exact sequence in homology.” So this trick tends to be useful in actually calculations of your derived functor. I won’t go through it, since it is just your standard “Snake Lemma” construction.

When I said there was extra structure, I was thinking about going into putting a product on the whole thing to make it into a graded ring, but I’ve decided that that is getting a little far afield for now. This may be the end of my ramblings on derived things for awhile.

The other thing I thought I should mention was my confusion on what this blog has become. I’m at a sort of turning point. I’m not sure if I should eliminate the non-math/mathematical physics and turn it into a blog just on that stuff (it sort of accidentally has shifted to that unofficially), or if I should make a conscious effort to balance things more. There are positives and negatives to both in my mind. Actually changing to a more focused blog would help draw and keep readers that actually care about the stuff I’ve been doing recently. On the other hand, it sort of goes against everything I believe in. But how its been going now, I’ve probably alienated all readers that used to read for philosophy or random art things, and so randomly posting on those things seems sort of pointless if I’ve lost those people and it will just serve to confuse and possibly alienate people only interested in the math side.

No immediate decisions will be made, so there is some time.

Follow

Get every new post delivered to your Inbox.

Join 206 other followers