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poincare duality notes notes reformatted, pictures (compare)

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• Mar 24 2020 01:08
The remaining task is to map to the separate joins.
Chinmaya Kausik
@Chinmaya-Kausik

The triple product:

• We start with three spaces $X$, $Y$ and $Z$.
• We construct a quotient of $X\times Y\times Z \times \Delta$, with $\Delta$ a $2$-simplex.
• We take a quotient where at each vertex two of the three spaces are identified to points.

This is probably the formal linear combination approach you mentioned.

Added double  for latex

I see
Here we take a point in the triangle $\Delta$ and map the fibre over each point to a join.
By the way I am working this out as I type.
Let us call the vertices $X$, $Y$ and $Z$ based on which space is not identified to a point at a vertex.
First let us consider $(X * Y) * Z)$.
Clearly the edge from $X$ to $Y$ should be mapped to $X * Y$, and the map is an obvious one.
In general, any point in $\Delta$ other than $Z$ is on a line $l$\$ that is parallel to $XY$. Identify $l$ with $XY$ by projection from $Z$ to give an identification of the fibre over $l$ with $X * Y$.
I now see a mistake.
We also need to quotient along the sides, with the space $Z$ pinched to a point on the line $XY$ for example. Otherwise "Clearly the edge from XX to YY should be mapped to X * YX∗Y, and the map is an obvious one." has a problem - there may be a map but it is not a homeomorphism.
Now proceeding with this correction, we get $((x, y), s)$ by projecting radially from $Z$, and we get $t$ as the distance from the base $XY$, normalized so that $Z$ is at distance $1$.
The above is Wrong too.
At any point in the interior, $x$, $y$ and $z$ are all given. All I should say is that we get $s$ by projecting radially from ZZ, and we get tt as the distance from the base XYXY, normalized so that ZZ is at distance $1$.
What is to be verified is this being well-defined at $Z$ and along the sides $XZ$ and $YZ$.
I now see what is happening.
• A priori we have a map on
KaTeX parse error: Undefined control sequence: \Y at position 9: X \times\̲Y̲ ̲\times Z\times …: X \times\Y \times Z\times R
where $R = [0,1]\times [0, 1]$
• At $t = 1$ the value of $s$ "does not matter" in $((X * Y) * Z)$, as $X * Y$ has been collapsed to a point.
• This allows one to pass to the quotient of the rectangle, where the top edge is pinched. This edge is just a triangle.
I now see what is happening.
• A priori we have a map on $X \times Y \times Z\times R$ where $R = [0,1]\times [0, 1]$
• At $t = 1$ the value of $s$ "does not matter" in $((X * Y) * Z)$, as $X * Y$ has been collapsed to a point.
• This allows one to pass to the quotient of the rectangle, where the top edge is pinched. This edge is just a triangle.
What I have described, starting with a triple product, may be the cleanest way to formalize this.
If the above is correct, then to complete the proof we map to $X * (Y * Z)$ by projecting from $X$ (instead of $Z$) to get $t$ (instead of $s$). Further, $s = 0$ at $X$ and $s = 1$ on $YZ$.
Chinmaya Kausik
@Chinmaya-Kausik
I see. So I guess the key insight is converting a rectangle to a simplex by passing to a quotient.
To achieve the formal linear combination picture.
That's what it looks like. Indeed one converts a rectangle into a simplex in two different ways for the two different associations. Once this is done, the two coincide.
Chinmaya Kausik
@Chinmaya-Kausik
I see, thank you so much.
Thank You. I too understand this clearly now.
If any of you is online can you go to teams.microsoft.com and log in with your iisc account. I want to test this as an alternative, since we have a paid subscription.
Chinmaya Kausik
@Chinmaya-Kausik
As you may have seen on teams.
AR-MA210
@AR-MA210
I saw another version of Whitehead's theorem somewhere, for homology groups, stated as "If a continuous map between simply connected CW complexes $f:X \to Y$ induces isomorphisms $H_*(X) \cong H_*(Y)$, $f$ is a homotopy equivalence.". Is it possible to prove this using the theorem from today's class + Hurewicz's theorem or some other tools we have?
It is another theorem.
It is in Hatcher. It follows from the result that I stated and did not prove, plus Hurewicz theorem and a little more
AR-MA210
@AR-MA210
Ok, I'll look into it. Thanks
I have started the teams, but not yet the video.
Sumanta
@dasumanta
Okay.
When I went to the link, I found they even have a debian linux app (they will surely have windows and mac), so I downloaded this. It may be good for everyone to do the same.
Shabarish CH
I am on the app for MacOS
philipthomasarun
@philipthomasarun
I am on the app on Windows
Chinmaya Kausik
@Chinmaya-Kausik
I'm trying to launch it on Chrome
Shabarish CH
It told me to restart the app to join the meeting
Chinmaya Kausik
@Chinmaya-Kausik
I had a question for Prof. Gadgil ( @siddhartha-gadgil ). We have seen lens spaces, but I think we're not sure why studying them is useful/relevant/interesting. I thought you would have some important insights on this, also because it seems you have worked on them in the past.
There are many interesting things with them. First of all, they are the historically first, and still prototypical examples for homotopy equivalence not coiniding with homeomorphism.
To show this involves both homotopy classification and homeomorphism classification. Both involve interesting ideas.
They are also geometric manifolds.
Further, Milnor's "Two complexes that are homeomorphic but combinatorially distinct" are derived from lens spaces.
Chinmaya Kausik
@Chinmaya-Kausik

To show this involves both homotopy classification and homeomorphism classification. Both involve interesting ideas.

I see. Is there some sort of punchline here? Or are the ideas themselves somewhat non-trivial?

Let me stick to $L(p, q)$ with $p$ prime. Firstly, any isomorphism of the fundamental group is induced by a map, but its degree $(mod p)$ is determined by the isomorphism.