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I'm trying to get my hands on the general spin group $G = \textrm{GSpin}_{2n}$. It have seen it mentioned as a connected, reductive group whose derived group is $\textrm{Spin}_{2n}$, which is the unique semisimple, simply connected group having the same root system as $\textrm{SO}_{2n}$.

The general spin group is for example mentioned in a paper here https://math.okstate.edu/people/asgari/Files/gspin.pdf .

I have read that it is difficult to view $\textrm{Spin}_{2n}$ as a closed subgroup of some general linear group. According to (9.16) in Linear Algebraic Groups and Finite Groups of Lie Type, the smallest $m$ for which $\textrm{Spin}_{2n}$ can be embedded in $\textrm{GL}_{m}$ is $2^{[\frac{2n-1}{2}]}$.

As for $G$, I am not really sure how that should be defined. Since its derived group is $\textrm{Spin}_{2n}$, it should be quotient of $\textrm{Spin}_{2n} \times S$ by a finite normal subgroup, where $S$ is some torus.

So my questions are:

1 . How should the general spin group be defined?

2 . What is the most straightforward way to compute the roots, coroots, and root datum of the general spin group as well as its derived group?

For the group $\textrm{SO}_{2n}$, I did (2) by finding a maximal torus and computing the Lie algebra. I imagine there must be a different approach when one is not working with an explicit embedding into $\textrm{GL}_{m}$.

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    $\begingroup$ By the definition in the cited paper, the connected dual Langland group to ${\rm GSpin}_{2n}$ is ${\rm GSO}_{2n}$. You find a definition of ${\rm GSO}_{2n}$ , say, in this thesis and compute its root datum. Then the dual root datum is the root datum of ${\rm GSpin}_{2n}$. $\endgroup$ – Mikhail Borovoi Nov 26 '16 at 15:55
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    $\begingroup$ My previous comment answers your second question. What about your first question, the group ${\rm GSpin}_{2n}$ is defined in the cited paper in terms of its connected dual Langlands group. In other words, you know its root datum, see my previous comment. $\endgroup$ – Mikhail Borovoi Nov 26 '16 at 16:03
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    $\begingroup$ Is there a need to consider only even orthogonal groups here? The group SO$_{2n+1}$ and its simply connected covering group Spin${2n+1}$ (of Lie type $B_n$) are comparable in many respects to the groups of type $D_n$. . $\endgroup$ – Jim Humphreys Nov 26 '16 at 20:58
  • $\begingroup$ P.S. To correct my typo, it should read Spin$_{2n+1}$ in the second line. $\endgroup$ – Jim Humphreys Nov 27 '16 at 14:44
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A good concise reference is Deligne's article on the Weil conjectures for K3 surfaces. See "La conjecture de Weil pour les surfaces K3" in Inventiones 15 (1971/72): 206-226. An English version is easy to Google too. In Deligne, and in other sources, the group GSpin is called CSpin. Here are the pertinent details.

Let $(V,q)$ be a finite-dimensional vector space with nondegenerate quadratic form, over any field (any base scheme is fine, adapting definitions appropriately... see the comment by nfdc23 below). Let $C(V,q)$ be the Clifford algebra, and $C^+$ the even part of the Clifford algebra. Embed $V$ in $C(V,q)$ as the degree 1 part; thus for $v \in V$, we view $v \in C(V,q)$ and $v \cdot v = q(v)$.

Then the group $CSpin$ is defined: $$CSpin(V,q) = \{ g \in C^+ : g V g^{-1} = V \}.$$ From above, we have a natural homomorphism $CSpin(V,q) \rightarrow SO(V,q)$ sending $g \in CSpin(V,q)$ to the map $(v \mapsto g v g^{-1})$. The kernel consists of scalars, giving the short exact sequence $$1 \rightarrow G_m \rightarrow CSpin(V,q) \rightarrow SO(V,q) \rightarrow 1.$$

The group $Spin(V,q)$ is the kernel of the spinor norm $CSpin(V,q) \rightarrow G_m$. Altogether, this gives the diagram used by Deligne: $$ \begin{array}{ccccc} & & Spin(V,q) & & \\ & & \downarrow & \searrow & \\ G_m & \rightarrow & CSpin(V,q) & \rightarrow & SO(V,q) \\ & \searrow & \downarrow & & \\ & & G_m & & \\ \end{array} $$ (Excuse the pooly formatted diagram!) The bottom-left diagonal arrow is the map $x \mapsto x^{-2}$.

This pretty quickly gives an identification $$CSpin(V,q) \cong \frac{Spin(V,q) \times G_m}{\mu_2}.$$ Here $\mu_2$ is identified with the kernel of $Spin(V,q) \rightarrow SO(V,q)$.

If you want to describe this $\mu_2$ using coroots, the only subtlety is in type $D_n$ with $n$ even, i.e., in $Spin_{4n}$. (Thanks to Mikhail Borovoi for the correction and nfdc23 for the suggestion.) Outside of $n=4$, the central $\mu_2$ in $Spin(V,q)$ should be the only one fixed by the Dynkin diagram automorphism. In type $D_4$, i.e., for $Spin_8$, the representation $Spin(V,q) \rightarrow SO(V,q)$ determines a node on the Dynkin diagram. The central $\mu_2$ in $Spin(V,q)$ is the only one fixed by the Dynkin diagram automorphism fixing that node.

This makes it pretty easy to relate the root datum of a (split) $CSpin$ group to that of the simply-connected $Spin$ group (which is its derived subgroup).

For simply-connected groups, one can look up the full root datum in any good reference (e.g., Bourbaki).

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    $\begingroup$ You probably mean: "the only subtlety is in type $D_n$ with $n$ even". Indeed, when $n$ is odd the center of ${\rm Spin}_{2n}$ is isomorphic to $\mu_4$, so it has only one subgroup isomorphic to $\mu_2$. $\endgroup$ – Mikhail Borovoi Nov 27 '16 at 11:47
  • $\begingroup$ Marty, the construction and properties in your answer work over fields of all characteristics (even 2), and also work over any scheme with quadratic forms on vector bundles (where "non-degenerate" means that the quadratic form is fiberwise nonzero over the base and the associated projective quadric is smooth over the base). You might want to mention the link between the root system and the rank $d$ of $V$ (i.e., type B$_n$ if $d=2n+1 \ge 3$ and type D$_n$ if $d=2n \ge 4$). $\endgroup$ – nfdc23 Nov 27 '16 at 15:05
  • $\begingroup$ I've made edits based on the two comments above. $\endgroup$ – Marty Nov 27 '16 at 16:58
  • $\begingroup$ Your CSpin is the question’s GSpin, right? $\endgroup$ – LSpice Dec 8 '19 at 3:24
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If you'd like to be completely explicit, the Atlas of Lie Groups and Representations software can do this. See www.liegroups.org. Comments are in braces { }.

atlas> {start with simply connected group Spin(8)xGL(1),
of type D4.T1, then mod out by specified element of order 2
so G=Spin(8)xGL(1)/Z_2
Z_2 given by the vector [1/2,1/2,1/2]: element of center
of Spin(8)xGL(1)=Z/2 x Z/2 x C^x, i.e. the element (-1,-1,-1)}
atlas> set rd=root_datum(Lie_type ("D4.T1"),[1/2,1/2,1/2])
Variable rd: RootDatum
atlas> rd
Value: simply connected root datum of Lie type 'D4.T1'
{derived group is simply connected}
atlas> set roots=simple_roots (rd)  
{simple roots as columns of a matrix}
Variable roots: mat
{The software chooses a basis so that X^*(T)=Z^5, X_*(T)=Z^5, 
with the dot product pairing, and then gives roots and corooots 
as integral vectors}
atlas> set coroots=simple_coroots (rd)  {simple coroots as columns}
Variable coroots: mat
atlas> roots
Value:  {roots are columns, 4 vectors of size 5} 
|  2, -1,  0,  0 |
|  3,  0, -1, -1 |
|  1,  0,  0,  0 |
| -2,  0,  0,  2 |
|  0,  0,  0,  0 |

atlas> coroots
Value: {coroots are columns}
| 1, -2,  1, 1 |
| 0,  1, -2, 0 |
| 0,  0,  2, 0 |
| 0,  0, -1, 1 | 
| 0,  0, -1, 0 |

atlas> ^roots*coroots  {compute the pairing of roots and coroots}
Value: 
|  2, -1,  0,  0 |
| -1,  2, -1, -1 |
|  0, -1,  2,  0 |
|  0, -1,  0,  2 |

atlas> ^roots*coroots=Cartan_matrix(rd)
Value: true  {this confirms that these are valid roots and 
coroots of the root system}
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