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Chinmaya Kausik
@Chinmaya-Kausik
To define an initial topology
Siddhartha Gadgil
@siddhartha-gadgil

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.

The remaining task is to map to the separate joins.
Chinmaya Kausik
@Chinmaya-Kausik

The triple product:

  • We start with three spaces XX, YY and ZZ.
  • We construct a quotient of X×Y×Z×ΔX\times Y\times Z \times \Delta, with Δ\Delta a 22-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
Siddhartha Gadgil
@siddhartha-gadgil
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 XX, YY and ZZ based on which space is not identified to a point at a vertex.
First let us consider (XY)Z)(X * Y) * Z).
Clearly the edge from XX to YY should be mapped to XYX * Y, and the map is an obvious one.
Siddhartha Gadgil
@siddhartha-gadgil
In general, any point in Δ\Delta other than ZZ is on a line ll$ that is parallel to XYXY. Identify ll with XYXY by projection from ZZ to give an identification of the fibre over ll with XYX * Y.
I now see a mistake.
We also need to quotient along the sides, with the space ZZ pinched to a point on the line XYXY 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)((x, y), s) by projecting radially from ZZ, and we get tt as the distance from the base XYXY, normalized so that ZZ is at distance 11.
The above is Wrong too.
Siddhartha Gadgil
@siddhartha-gadgil
At any point in the interior, xx, yy and zz are all given. All I should say is that we get ss by projecting radially from ZZ, and we get tt as the distance from the base XYXY, normalized so that ZZ is at distance 11.
What is to be verified is this being well-defined at ZZ and along the sides XZXZ and YZYZ.
Siddhartha Gadgil
@siddhartha-gadgil
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]×[0,1]R = [0,1]\times [0, 1]
  • At t=1t = 1 the value of ss "does not matter" in ((XY)Z)((X * Y) * Z), as XYX * 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×Y×Z×RX \times Y \times Z\times R where R=[0,1]×[0,1]R = [0,1]\times [0, 1]
  • At t=1t = 1 the value of ss "does not matter" in ((XY)Z)((X * Y) * Z), as XYX * 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(YZ)X * (Y * Z) by projecting from XX (instead of ZZ) to get tt (instead of ss). Further, s=0s = 0 at XX and s=1s = 1 on YZYZ.
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.
Siddhartha Gadgil
@siddhartha-gadgil
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.
Siddhartha Gadgil
@siddhartha-gadgil
Thank You. I too understand this clearly now.
Siddhartha Gadgil
@siddhartha-gadgil
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
Oh, I see. Hadn't read this. I am online.
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:XYf:X \to Y induces isomorphisms H(X)H(Y)H_*(X) \cong H_*(Y), ff 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?
Siddhartha Gadgil
@siddhartha-gadgil
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
Siddhartha Gadgil
@siddhartha-gadgil
I have started the teams, but not yet the video.
Sumanta
@dasumanta
Okay.
Siddhartha Gadgil
@siddhartha-gadgil
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
@shabarishch_twitter
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
@shabarishch_twitter
It told me to restart the app to join the meeting
I did get the notification
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.
Siddhartha Gadgil
@siddhartha-gadgil
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?

Siddhartha Gadgil
@siddhartha-gadgil
I would say more than one punchline.
Let me stick to L(p,q)L(p, q) with pp prime. Firstly, any isomorphism of the fundamental group is induced by a map, but its degree (modp)(mod p) is determined by the isomorphism.
Hence whether this is a homotopy equivalence, and further orientation preserving, can be determined.