Difference between revisions of "A Portal Special Presentation- Geometric Unity: A First Look"

''[https://youtu.be/Z7rd04KzLcg?t=6774 01:52:54]''<br>

''[https://youtu.be/Z7rd04KzLcg?t=6774 01:52:54]''<br>

And when you linearize that, if you are in low enough dimension, you have \(\Omega^0\), \(\Omega^1\), sometimes \(\Omega^0\) again, and then something that comes from \(\Omega^2\), and if you can get that sequence to terminate by looking at something like a half-signature theorem or a bent back de Rham complex in the case of dimension three, you have an Atiyah-Singer theory. And remember we need some way to get out of infinite dimensional trouble, right? You have to have someone to call when things go wrong overseas and you have to be able to get your way home. And in some sense, we call on Atiyah and Singer and say we're in some infinite dimensional space, can't you cut out some finite dimensional problem that we can solve even though we start getting ourselves into serious trouble? And so we're going to do the same thing down here. We're going to have \(\Omega^0(ad)\), \(\Omega^1(ad)\), direct sum \(\Omega^0(ad)\), \(\Omega^{d−1}(ad)\), and it's almost the same operators.

And when you linearize that, if you are in low enough dimension, you have \(\Omega^0\), \(\Omega^1\), sometimes \(\Omega^0\) again, and then something that comes from \(\Omega^2\), and if you can get that sequence to terminate by looking at something like a half-signature theorem or a bent back de Rham complex in the case of dimension three, you have an Atiyah-Singer theory. And remember we need some way to get out of infinite dimensional trouble, right? You have to have someone to call when things go wrong overseas and you have to be able to get your way home. And in some sense, we call on Atiyah and Singer and say we're in some infinite dimensional space, can't you cut out some finite dimensional problem that we can solve even though we start getting ourselves into serious trouble? And so we're going to do the same thing down here. We're going to have \(\Omega^0(ad)\), \(\Omega^1(ad)\), direct sum \(\Omega^0(ad)\), \(\Omega^{d−1}(ad)\), and it's almost the same operators.

''[https://youtu.be/Z7rd04KzLcg?t=6849 01:54:09]''<br>

''[https://youtu.be/Z7rd04KzLcg?t=6849 01:54:09]''<br>