Can someone explain, explicitly, how to, given a reductive complex algebraic group construct the Langlands dual group? I know it is a group with the cocharacters of G as its characters, but how does one go about writing down what group it is?
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asked Oct 22, 2009 at 23:03
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$\begingroup$ Note that the word "complex" isn't needed here, just an algebraically closed field of arbitrary characteristic (as noted by Marc Palm). Springer's book is one source. $\endgroup$– Jim HumphreysCommented Nov 15, 2019 at 22:55
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7 Answers
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You can construct the dual group in a combinatorial manner as follows: Reductive groups are classified by their root datum. There is an obvious duality on the set of all root data, and the dual group is the reductive group with the dual root datum.
You can see Wikipedia for the notion of the root datum.
answered Oct 23, 2009 at 0:33
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3$\begingroup$ Thanks, I guess that's more-or-less what I wanted, is there a good source for the construction of the algebraic group from the root datum? $\endgroup$ Commented Oct 23, 2009 at 0:58
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1$\begingroup$ Yes, there's an explicit (infinite) presentation of the points over any field (ring?) due to Chevalley. It's more convenient to just use a representation, though. $\endgroup$– Ben Webster ♦Commented Dec 31, 2009 at 18:51
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$\begingroup$ @BenWebster, if you try using the Chevalley presentation over an arbitrary ring even in type $\mathsf A_n$, then you run into K-theoretic troubles! $\endgroup$– LSpiceCommented Nov 15, 2019 at 19:48
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The Chevalley construction mentioned by Steven Sam from Lusztig's notes can also be found in Springer's textbook "Linear algebraic groups" (on pages 164-165 of the second edition). It's beautiful and clear and elementary (no perverse sheaves!) and direct (no Lie algebras, which wouldn't work in finite characteristic, and no reduction to simply connected groups). If you want to construct representations of the dual group directly, then geometric Satake is a good idea. For the classical Langlands conjectures, what one cares about are not representations of the dual group but elements of it (or a little more precisely, maps of Galois groups into it). For such purposes, the Chevalley construction by generators and relations is just what Dr. Langlands ordered.
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One way to "construct" the Langlands dual group is via the geometric Satake isomorphism, which gives an equivalence between the category of representations of the Langlands dual group and the category of G[[t]]-equivariant perverse sheaves on the affine Grassmannian. From the category of representations, the group can be reconstructed via Tannakian duality, i.e., as automorphisms of the fibre functor. I suppose whether or not this is an explicit construction depends on your point of view.
answered Oct 23, 2009 at 0:10
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$\begingroup$ Is there a way to do it without perverse sheaves? I don't know anything about them (yet). $\endgroup$ Commented Oct 23, 2009 at 0:14
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$\begingroup$ I imagine there's a way to construct the representation category of a reductive group directly from the root data, but I'm not aware of the details (or anywhere where the details are written out in a nice way). Probably one first constructs the associated Lie algebra and considers a certain subcategory of the category of representations of the Lie algebra. From there, the Tannakian formalism gives a construction of the group. It'd be great if someone can provide a good reference for such a construction. James Milne appears to be writing one, but it's not yet complete. $\endgroup$ Commented Oct 23, 2009 at 0:26
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1$\begingroup$ Also, another point of view of the geometric Satake isomorphism is that it gives an equivalence between the category of representations of the Langlands dual group and the category of "Hecke functors" on the moduli stack of G-bundles on an algebraic curve. A brief introduction to these things is given at the beginning of the first lecture of Dennis Gaitsgory's graduate seminar (math.harvard.edu/~gaitsgde/grad_2009). $\endgroup$ Commented Oct 23, 2009 at 0:39
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$\begingroup$ Reply by Sean Rostami (who didn't have enough rep to comment): I'm not sure if this is what you mean, but my gut reaction is to use the highest weight theorem and various formulas (e.g. Kostant). You index the irreducible representations by the dominant characters and use the formulas to figure out the dimensions of all the weight spaces and how to write tensor products as direct sums of irreducibles. $\endgroup$ Commented Dec 29, 2009 at 21:03
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1$\begingroup$ One advantage of geometric Satake is that it gives a construction over arbitrary commutative rings, including those for which the representation category is not semisimple. There is a variant using Whittaker sheaves that yields a derived equivalence as well (the Satake equivariant derived category is not as well-behaved). $\endgroup$– S. Carnahan ♦Commented Jan 4, 2010 at 4:27
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Lusztig explained how to construct a reductive algebraic group from a given root datum in a class I took from him. I took notes and scanned them. If you don't mind the quality, you can find the construction in Lecture 14 (starting from p.34) here: http://math.mit.edu/~ssam/notes/alggp.djvu
The construction is similar to constructing a semisimple Lie algebra from a root system.
answered Oct 23, 2009 at 2:49
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$\begingroup$ Is there an updated link? Link is broken. Thanks :) $\endgroup$– FNHCommented Dec 2, 2023 at 20:59
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As Peter mentioned, reductive groups are determined by their root data, and the Langlands dual is given by switching weights with coweights, and roots with coroots.
There is a "construction" of a group from a root datum in SGA III Exp 25 (in vol 3). It starts by reducing to the simply connected semisimple case (meaning there are ways of going from this case to the general case). The weights give you a torus $T$, and the positive/negative roots give you unipotent groups $U^+$ and $U^-$. You form a scheme $\Omega = U^- \times T \times U^+$, and create $G$ by gluing a few disjoint copies of $\Omega$ together, and writing down a composition law. This yields a split group over an arbitrary base scheme.
answered Oct 23, 2009 at 2:54
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I imagine there's a way to construct the representation category of a reductive group directly from the root data, but I'm not aware of the details (or anywhere where the details are written out in a nice way). [...] – Evan Jenkins Oct 23 at 0:26
Again, I'm always responding to old posts... (do I need to keep mentioning this or is it perfectly acceptable to reply to old questions?). I'm not sure if this is what you mean, but my gut reaction is to use the highest weight theorem and various formulas (e.g. Kostant). You index the irreducible representations by the dominant characters and use the formulas to figure out the dimensions of all the weight spaces and how to write tensor products as direct sums of irreducibles.
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1$\begingroup$ The conversation never stops on MO, as better or more detailed answers and comments can be posted at any time, so you're welcome to post anything that benefits the post. (The only constraint, I would say, is to look at what others posted before jumping with an answer) By the way, since here you're responding to the comment, I posted it to the place it belongs above. You should be able to comment yourself pretty soon. $\endgroup$ Commented Dec 29, 2009 at 21:07
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I was googling for the root datum. This site and a very nice, conscise treatment of Calder Daenzer turned up: http://www.math.psu.edu/daenzer/Root_Data_for_Reductive_Groups.pdf His notes deal with complex, reductive groups only (as requested by the OP). In the end, he states the definition of the Langlands dual group.
Perhaps, you will also like Casselman: http://www.math.ubc.ca/~gor/Notes.pdf
I am aware that this question is pretty old! I add this as answer because I was not quite satisfied with the remaining suggestions in relation of brevity/precision. E.g. Springer's book requires a massive book keeping of notation. I am aware that Springer's/the SGA proof works over any basefield, and is thus more general.
answered Apr 23, 2013 at 15:21