Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be algebraic group over $\mathbb(C)$(semisimple), $B$ be Borel subgroup. consider flag variety $G/B$. It is known that $G/B$ is projective variety. So one consider the homogeneous coordinate ring $ \mathbb{C}[G/B]$, it is known that it is a graded polynomial ring quotient by some homogeneous ideal. For example, consider flag variety of $sl_2$, it is $\mathbb{P}^{1}$, so the coordinate ring is $\mathbb{C}[x_1,x_2]$ as graded ring.

I want to know for general case, i.e. flag variety of finite dimensional Lie algebra $g$($G/B$), how to compute $\mathbb{C}[G/B]$? But we know that from Borel-Weil, we can write it as

$\bigoplus_{\lambda\in P_+}$ $R_\lambda$, where $R_\lambda$ is irreducible highest dominant weight representations of $g$

My question

How to compute $\mathbb{C}[G/B]$? explicitly?

How to build explicit ring isomorphism from $\bigoplus_{\lambda\in P_+}$ $R_\lambda$ to some concrete ring?

This question might be elementary. Thanks anonymous and Ben pointing out the stupid mistake I made

share|improve this question
Sorry, I should think a little more. Now I figured out for $sl_2$ case, just identify the extreme vector with $x_1$ or $x_2$. Then the monomial in homogeneous component of coordinate ring $\mathbb{C}[x_1,x_2]$ can correspondence to some $R_\lambda$ –  Shizhuo Zhang May 4 '10 at 14:19
Shizhuo- If the homogeneous coordinate ring of a variety is a polynomial ring, then that variety is a weighted $\mathbb{P}^n$, so this is only true in $\mathfrak{sl}_2$. –  Ben Webster May 4 '10 at 14:33
Sorry for making stupid mistake, I have modified this question –  Shizhuo Zhang May 4 '10 at 14:50

3 Answers 3

up vote 6 down vote accepted

$G/B$ is most naturally a multi-projective variety, embedding in the product of projectivizations of fundamental representations: $\prod \mathbb{P} (R_{\omega_i})$. So there is a multi-homogeneous coordinated ring on $G/B$. You mentioned that this ring is $\bigoplus_{\lambda\in P_+}$ $R_\lambda$. This is correct, and the grading is also apparent. It's given by the weight lattice. (To be more canonical, you should take the duals of every highest weight representation, but what you've written down is isomorphic to that.)

So all that remains is giving the multiplication law on $\bigoplus_{\lambda\in P_+}$ $R_\lambda$. You need to specify maps $R_\lambda \otimes R_\mu \mapsto R_{\lambda+\mu}$. There's a natural candidate: If you decompose $R_\lambda \otimes R_\mu$ into a direct sum of irreducible representations, $R_{\lambda+\mu}$ will appear exactly once. The multiplication law is simply projection onto this factor.

@Shizuo: For $sl_n$, the situation is more explicit. The flag variety here is the set of flags $0 = V_0 \subset V_1 \cdots V_{n-1} \subset V_n = \mathbb{C}^n$ with $\mathrm{dim} V_i = i$. So the flag variety is a closed subvariety of the product of Grassmannians $Gr(1,n) \times \cdots Gr(n,n) $. Each of these Grassmannians have a explicitly Plucker embedding into the projectivization of the exterior power of $\mathbb{C}^n$. In particular, the homogeneous ideal is explicitly given by the Plucker relations. So the multi-homogeneous coordinate ring of $Gr(1,n) \times \cdots Gr(n,n) $ is just the tensor product of the known homogeneous coordinate rings.

Finally, to get the multi-homogeneous coordinate ring for the flag variety, we need to specify an incidence locus inside $Gr(1,n) \times \cdots Gr(n,n) $. Namely, we need to specify those tuples of subspaces that form a flag. But this is easy: it corresponds to certain wedge products being zero. Just impose those additional relations, i.e. mod out by the corresponding multi-homogenous ideal. Now you should have an explicit, albeit fairly long description of the homogeneous coordinate ring. The above answer for $SL_3$ looks like a special case of this construction.

share|improve this answer
Thanks. Yes, I know this multiplication structure given by projection. But how to compute them. How to compute some non-trivial examples such as sl_n –  Shizhuo Zhang May 4 '10 at 15:47
A side note: It is not literally true that the Plucker relations simply say that things wedge to zero, although this is an excellent intuition for what they say. For example, in $SL_5$, consider a point $(V_1, V_2, V_3, V_4) \in Gr(1,5) \times Gr(2,5) \times Gr(3,5) \times G(4,5)$ such that $V_1$ is contained in all the $V_i$, $V_4$ contains all the $V_i$, but $V_3$ does not contain $V_2$. I think you will find that every wedge product which vanishes on $Fl_5$ is also zero here. (Note that $V_2 \wedge V_3$ is zero because they have a one dimensional intersection.) –  David Speyer May 4 '10 at 17:30
What about when $G$ is Lie group and we take $G/B$ as flag variety? –  Hassan Jolany Mar 15 at 20:00

The ring in question may be identified with the ring of $N$-invariant functions on $G$. (Where $N$ is the unipotent radical of $G$.)

In type $A$, the group $G$ is $n \times n$ matrices with determinant $1$. A function is $N$ invariant if it is preserved by rightward column operations. (That is, adding $a \times (\mbox{column} \ i)$ to $\mbox{column} \ j$, for $i<j$.) It is obvious that those minors which use the left $k$ columns for some $k$ have this invariance.

In fact the ring is generated by these minors, which are called Plucker coordinates. The relations between these minors, called Plucker relations, are also extremely classical and well known. A good low level reference to this topic is Chapter 14 of Miller and Sturmfels' book Combinatorial Commutative Algebra.

share|improve this answer
Thank you very much, actually, I just want to get myself comfortable with some non-trivial coordinate ring calculation. –  Shizhuo Zhang May 4 '10 at 18:20
A couple more references where this is computed very explicitly: Roger Howe, Perspectives on invariant theory: Schur duality, multiplicity-free actions and beyond. Schur lectures (long survey) Claudio Procesi, Lie groups. An approach through invariants and representations (textbook) –  Victor Protsak May 4 '10 at 19:09

The ring $\bigoplus R_{\lambda}$ is generally not a polynomial ring. For example, when $G=SL_3$, this ring is $$\mathbb{C}[p_1, p_2, p_3, p_{12}, p_{13}, p_{23}]/(p_1 p_{23} - p_{13} p_2 + p_{3} p_{12}).$$

In the case of $SL_2$, this is simple enough. The group $SL_2$ acts on the vector space $V$ spanned by $x$ and $y$ in the obvious way. The homogenous degree $n$ polynomials are naturally $\mathrm{Sym}^n \ V$ and thus acquire an action of $SL_2$ as well. One sees that this is the irreducible representation of degree $n+1$.

share|improve this answer
can you give me a reference to tell how to compute this? –  Shizhuo Zhang May 4 '10 at 14:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.