WEBVTT
01:36:51.000 --> 01:37:01.000
and
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I I I think that there probably is some connection between pie and pie.
01:37:10.000 --> 01:37:18.000
Check in in this respect, but it isn't simple as Jeff said.
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It can easily happen that one of them has an irreducible, associated variety in the other.
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Does not but I think there's a connection and that there's good manager in understanding the nature of that connection.
01:37:36.000 --> 01:37:55.000
So i'll i'll just mention one or theorem, if the boundary of this complex orbit oh, oh, check!
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Oh, well, they are group side. If the boundary has coda mentioned at least 4 in in the closure of all
01:38:14.000 --> 01:38:20.000
So that code in that code dimension is always even so so it's 2, 4, 6, or something, if it's at least 4.
01:38:20.000 --> 01:38:33.000
Then the associated variety of pi must be irreducible
01:38:33.000 --> 01:38:46.000
So there there's some interesting geometry going on here
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Anyway.
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Weekly, fair range or something. The assessment variety for an Aq.
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Lambda is always a reducible in yeah weekly fair
01:39:36.000 --> 01:39:41.000
Sorry the statement of the theorem is if you're in the Weekly Fair Range.
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Then the associated variety is therereducible. Yeah.
01:39:45.000 --> 01:39:54.000
Sorry if if Pi is the aqlanda is in the weekly fair range that's an assumption.
01:39:54.000 --> 01:40:01.000
Then, the associated variety of Pi. is there your soul?
01:40:01.000 --> 01:40:11.000
Thank you. and if Pi is in the same cell as some weekly fair, A. Q.
01:40:11.000 --> 01:40:21.000
And a the same conclusion, because the associated variety is constant on the sales.
01:40:21.000 --> 01:40:33.000
So that gives you lots and lots of associated varieties which are reusable. and of course, a lot of the interesting examples that we look at involving important representations is where this can fail.
01:40:33.000 --> 01:40:41.000
But questions.
01:40:41.000 --> 01:40:46.000
Okay. So I was thinking I could do a couple of examples.
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To do
01:40:52.000 --> 01:40:57.000
To do what I want to say. Well, so let's just go to the software.
01:40:57.000 --> 01:41:03.000
Do a couple of examples. So maybe this should be larger.
01:41:03.000 --> 01:41:14.000
Yeah, ,
01:41:14.000 --> 01:41:30.000
That's fine that's fine so let's let's set g is equal to and so 2
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Is the week. Yes, Gee!
01:41:39.000 --> 01:41:44.000
So here are the unipolent parameters for test.
01:41:44.000 --> 01:42:02.000
P. S. L. 2 R. so you know, let me let me cut and paste this into the other window, so
01:42:02.000 --> 01:42:16.000
Oops! right
01:42:16.000 --> 01:42:24.000
Yeah.
01:42:24.000 --> 01:42:38.000
Okay, so So, for example, here's the the orbit on the dual side is the 0 orbit.
01:42:38.000 --> 01:42:46.000
And this is an example I talked about last time. These are the 3 representations with infinitesimal characters. 0.
01:42:46.000 --> 01:42:53.000
There's 2 limits of discrete series
01:42:53.000 --> 01:43:14.000
And there's. an irreducible spherical principal series, and the only other one is when you take the the regular orbit, there is just the trivial representation
01:43:14.000 --> 01:43:26.000
And in terms of honest arthur packets I didn't think about it.
01:43:26.000 --> 01:43:30.000
Do more than one
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Is there more than one? I think so, because the other dual side G check is Pso.
01:43:37.000 --> 01:43:41.000
2,
01:43:41.000 --> 01:43:55.000
And so, if you take the principal orbit it's centralizer is trivial
01:43:55.000 --> 01:44:03.000
There's only one packet, I think let's do a more interesting example.
01:44:03.000 --> 01:44:33.000
I mean these are all of course let's look at G, 2
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Heard roll.
01:44:56.000 --> 01:45:09.000
Okay. So this represent down the dual side there there are 5 orbits
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Orbit to 0, 1, 2, 4, and then this interesting orbit 3.
01:45:17.000 --> 01:45:25.000
So this is some regular orbit, and well, so excellent exercise the trick.
01:45:25.000 --> 01:45:50.000
So the there are 4. These are the 4 week packets consisting of 2, 2, 2, 5, and one representation resp. 1, 2, plus 2, plus 2 plus 5, and an extremely interesting exercise
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You just figure out the honest Arthur packets
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In in each case, especially or number 3
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Questions.
01:46:28.000 --> 01:46:36.000
Do you have any recollection of about what's the honest packets are.
01:46:36.000 --> 01:46:49.000
Yeah, no. And for orbit number 3. no, I mean have some.
01:46:49.000 --> 01:47:02.000
They ideas, but not not clear. Well, okay, so that's a good that's a good excuse, that's a good it's good opportunity to
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Website
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Okay, that's very weird. stop sharing new share this come on
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Sorry about that was
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Sorry about that, was struggling with time where's my browser.
01:47:42.000 --> 01:47:48.000
I got a browser somewhere there is. Okay. Try one more time.
01:47:48.000 --> 01:48:04.000
Sorry share screen this up one so let's go to alice here are the here's this list of uniform representations which I was talking about.
01:48:04.000 --> 01:48:09.000
So here's the split split type split G.
01:48:09.000 --> 01:48:20.000
2 and I think I wanna look at
01:48:20.000 --> 01:48:29.000
That's rather small, I realize. yeah so I think there's quite a bit of information here.
01:48:29.000 --> 01:48:36.000
So this this the centralizer of this orbit is not a 1 billion.
01:48:36.000 --> 01:48:44.000
It's s 3 there it is right there s 3 and so here
01:48:44.000 --> 01:48:54.000
I claim 5 is equal to 2, plus 3 so apparently I Convince myself at some point that there are 2 disjoint packets of size 2 and 3 resp.
01:48:54.000 --> 01:49:09.000
But I don't remember how I did that so you're previous. The The one note page showed that these 5 representations belong to 2 different cells.
01:49:09.000 --> 01:49:20.000
Oh, yes, and in in G, 2: Yeah. There were 3 in one cell and 2 in the other.
01:49:20.000 --> 01:49:25.000
Yeah, I think that's right here's Oh, yeah, okay.
01:49:25.000 --> 01:49:31.000
I just remembered Then I I I mean I think that goes a long way.
01:49:31.000 --> 01:49:42.000
But there's some there's this fact right here these stars here indicate that in an eat the dual of each cell contains an Aq.
01:49:42.000 --> 01:49:55.000
Lambda. And so these associated varieties are all irreducible, which forces those 2 packets to be disjoint
01:49:55.000 --> 01:50:16.000
Something. Let me let me say that
01:50:16.000 --> 01:50:32.000
Yeah. So I just wanted to write that in in this example the stars here indicate that that the dual cell contains an Aq.
01:50:32.000 --> 01:50:41.000
Lambda, so that implies all of the the the Arthur packets are disjoint
01:50:41.000 --> 01:50:48.000
And from that it's not hard to see that this 5 is equal to 2 plus 3.
01:50:48.000 --> 01:51:03.000
And I think anyway, there's so Let me just let me just write a concluding sentence here.
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Conclusion
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We can compute. we meeting atlas computes this these by a low check 2.
01:51:22.000 --> 01:51:41.000
It there's lots of information to inform the decomposition of 5 o'clock is the union of the Pi.
01:51:41.000 --> 01:51:50.000
Oh, sigh! eyes but in general
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We cannot thank you. The individual, bye, size.
01:52:04.000 --> 01:52:11.000
Questions.
01:52:11.000 --> 01:52:37.000
So Arthur's theory attaches to an Arthur Parameter, a finite group a of and for these 2 packets in G 2, the the Arthur groups our our for the first one hey?
01:52:37.000 --> 01:52:42.000
Of size S. 3
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And for the second one a size s 2 s one, meaning the the second honest Arthur Packet.
01:52:54.000 --> 01:53:01.000
Attached to the same subregular orbit. Oh, sorry.
01:53:01.000 --> 01:53:15.000
I was thinking of size as an arthur a fall Arthur program the the sigh corresponding to the 3.
01:53:15.000 --> 01:53:27.000
So so we're fixing O check is the subregular right, and the side corresponding to the 3 representations that that corresponds to this.
01:53:27.000 --> 01:53:42.000
A of sign is S. 3, which has 3 irreucible representations, and the 2 representations
01:53:42.000 --> 01:54:07.000
In that case a of Si is Z 2 which has 2 irreducible representations, and in this case the the our Arthur tells us there's a map from pi of side to representations of Aosi not
01:54:07.000 --> 01:54:15.000
necessarily reduceable ones; and in this case
01:54:15.000 --> 01:54:32.000
It's a bijection in these cases these 2 cases it's a bijection
01:54:32.000 --> 01:54:56.000
Unless there's further questions that should be placed to stop. so based on some of the chat questions, and where we are, I I thought that it would be a reasonable thing to talk more about duality next week emphasizing
01:54:56.000 --> 01:55:17.000
the non integral case. And and what happens there? Oh, boy, I got a look at this chat, so you see what that's about the well, the chatter used the phrase poison that discussion.
01:55:17.000 --> 01:55:23.000
But by mentioning non integral things and that's probably accurate.
01:55:23.000 --> 01:55:32.000
Actually I I sent that comment only to you in order to avoid
01:55:32.000 --> 01:55:45.000
Yeah, Okay, So Oh, So And you see the the Chat Jeff asking to paste the
01:55:45.000 --> 01:55:56.000
Oh, I i'm not sure. Some some are pasting the chat questions
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I don't think we have access to the chats after, so I I I will to just add them, put put it and put a note at the bottom of this file or something I will try to do.
01:56:10.000 --> 01:56:16.000
That. Yes, and I will try not to post the the non public chats.
01:56:16.000 --> 01:56:22.000
Yeah. So the the there's a question which is about this infinitesimal character.
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Being regular, and also being non integral. and David suggesting that we should take a little time to talk about that.
01:56:37.000 --> 01:57:03.000
The The the short answer is that For this you need to take a subgroup g check of gamma on the dual side, and for this you need to use the translation principle to go from
01:57:03.000 --> 01:57:09.000
Rugby to singular, and those are both excellent questions with non trivial answers.
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So
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Other questions.