LECTURE 18: CLASSIFICATION OF ABSTRACT CARTAN MATRICES / DYNKIN DIAGRAMS

PROFESSOR: VICTOR KAČ TRANSCRIBED BY: STEVEN SIVEK AND ALEXEY SPIRIDONOV

1. Examples of Dynkin diagrams

In the examples and exercises that follow, we will compute the Cartan matrices (as described in the previous lecture) for the indecomposable root systems that we have encountered earlier. We record these as elegant Dynkin diagrams, summarized in Figure 1.1. Later in the lecture, we will prove that these are actually all the possible indecomposable root systems. We also compute extended Dynkin diagrams specifically for the purposes of this proof.

E_{7} E_{6} F_{4} G^{2}  Ẽ_{7} Ẽ_{6} F̃_{4} G̃_{2} Ẽ_{8} D̃_{r} C̃_{r} B̃_{r} Ã_{r} E_{8} B_{r} A_{r} D_{r} C_{r}  (i) (h) (g) (f) (e) (d) (c) (b) (a)

Figure 1.1. Dynkin diagrams of all the indecomposable root systems, from top to bottom: A_{r}, B_{r}, C_{r}, D_{r}, E_{8}, E_{7}, E_{6}, F_{4}, G_{2}.

In the following examples, the rank of the root system is always denoted by r, and we get simple roots α_{1}, …, α_{r}. The lowest negative root, used in the extended Dynkin diagram is denoted α_{0}. Perhaps the most detailed explanation of how to compute a Cartan matrix is given in Exercise 1.

Example 1.1. A_{r}: (ε_{i}, ε_{j}) = δ_{ij}, Δ = Δ_{sl_{r+1}(𝔽)} = {ε_{i} − ε_{j}|i, j ∈ [r + 1]} ⊂ V, where V is the subspace of ⨁_{i=1}^{r+1} ℝε_{i}, on which the sum of coordinates (in the basis {ε_{i}}) is zero.

Take f ∈ V^{∗} given by f(ε_{1}) = r + 1, f(ε_{2}) = r, …, f(ε_{r+1}) = 1; hence, f ≠ 0 on Δ, and f is integer-valued on all the roots.

Then, Δ_{+} = {ε_{i} − ε_{j}|i < j}. What is Π? Clearly, α = ε_{i} − ε_{i+1} ∈ Π, since f(α) = 1. There are r simple roots altogether (since Π spans V) and r such α, so Π = {α_{i} = ε_{i} − ε_{i+1}}_{i=1}^{r}.

For the Cartan matrix, recall that:

A = (\frac{2(α_{i}, α_{j})}{(α_{i}, α_{i})})_{i,j=1}^{r} = ( \begin{matrix}2 & −1 & 0 & ⋯ & 0 \\ −1 & 2 & −1 & & ⋮ \\ 0 & −1 & 2 & ... & 0 \\ ⋮ & & ... & ... & −1 \\ 0 & ⋯ & 0 & −1 & 2 \end{matrix} ) .

Hence, the Dynkin diagram in Figure 1.1a.

The largest root is θ = ε_{1} − ε_{r+1}, so α_{0} = ε_{r+1} − ε_{1} (α_{0} denotes the lowest negative root). This yields the extended Dynkin diagram (from the matrix on roots α_{0}, α_{1} …, α_{r}, denoted Ã = (\frac{2(α_{i},α_{j})}{(α_{i},α_{i})})) in Figure 1.1a.

Example 1.2. B_{r}: Δ = Δ_{so_{2r+1}(𝔽)} = {±ε_{i} ± ε_{j}(i ≠ j), ±ε_{i}}_{i,j=1}^{r} ⊂ V = ⨁_{i=1}^{r} ℝε_{i}.

Take the f analogously to A_{r} (but we have r, rather than r + 1 roots now): f(ε_{1}) = r, …, f(ε_{r}) = 1.

Then, Δ_{+} = {ε_{i} + ε_{j}(i ≠ j), ε_{i} − ε_{j}(i < j), ε_{i}}, and f = 1 for ε_{i} − ε_{i+1} and ε_{r}. Hence Π = {ε_{i} − ε_{i+1} (i = 1, …, r − 1), ε_{r}}. The Cartan matrix is almost identical, except that A_{r,r−1} = \frac{2(α_{r},α_{r−1})}{(α_{r},α_{r})} = 2(α_{r}, α_{r−1}) = −2, so

A = ( \begin{matrix}2 & −1 & 0 & ⋯ & 0 \\ −1 & 2 & −1 & & ⋮ \\ 0 & −1 & 2 & ... & 0 \\ ⋮ & & ... & ... & −1 \\ 0 & ⋯ & 0 & −2 & 2 \end{matrix} ) .

The lowest root is α_{0} = −(ε_{1} + ε_{2}), so we get the Dynkin diagrams in Figure 1.1b.

Example 1.3. C_{r}: Δ = Δ_{sp_{2r}(𝔽)}. The roots are the same as above, except ±ε_{i} becomes ±2ε_{i}; we also take the same f. Then, f(ε_{i} −ε_{i+1}) = 1 for i = 1, 2, …, r−1 (label these simple roots α_{1}, …, α_{r−1}), and we need an rth simple root. We might only possibly get f = 2 by summing some of the preceding α_{i}. However, it’s impossible to obtain 2ε_{r} this way. Hence, Π = {ε_{i} − ε_{i+1} (i = 1, …, r − 1), 2ε_{r}}. The new Cartan integers are \frac{2(α_{r−1},α_{r})}{(α_{r−1},α_{r−1})} = −2, and \frac{2(α_{r},α_{r−1})}{(α_{r},α_{r})} = −1, which gives us:

A = ( \begin{matrix}2 & −1 & 0 & ⋯ & 0 \\ −1 & 2 & −1 & & ⋮ \\ 0 & −1 & 2 & ... & 0 \\ ⋮ & & ... & ... & −2 \\ 0 & ⋯ & 0 & −1 & 2 \end{matrix} ) ,

and the Dynkin diagram in Figure 1.1c. The highest root is θ = 2ε_{1}, so α_{0} = −2ε_{1}, and we get the extended Dynkin diagram in Figure 1.1c.

Example 1.4. D_{r}: Δ = Δ_{so_{2r}(𝔽)} = {±ε_{i} ± ε_{j}(i ≠ j)}_{i,j=1}^{r}. Define f ∈ V^{∗} by f(ε_{1}) = r − 1, …, f(ε_{r}) = 0.

Then Δ = {ε_{i} ± ε_{j}(i < j)}, and Π = {ε_{1} − ε_{2}, …, ε_{r−1} − ε_{r}, ε_{r−1} + ε_{r}} since we get f(α_{i}) = 1 for all i.

Compute

A = ( \begin{matrix}2 & −1 & 0 & … & … & 0 \\ −1 & 2 & −1 & & & ⋮ \\ 0 & −1 & ... & ... & 0 & 0 \\ ⋮ & & ... & 2 & −1 & −1 \\ ⋮ & & 0 & −1 & 2 & 0 \\ 0 & … & 0 & −1 & 0 & 2 \end{matrix} ) ,

so we have the Dynkin diagram in Figure 1.1d.

Then, θ = ε_{1} + ε_{2}, so α_{0} = −ε_{1} − ε_{2}, so we get the D̃_{r} in Figure 1.1d.

Example 1.5. E_{8}: Δ = Δ_{E_{8}} = \{±ε_{i} ± ε_{j}(i ≠ j), \frac{1}{2} (±ε_{1} ± ε_{2} ± ⋯ ± ε_{8})_{(even number of + signs)}\}_{i,j=1}^{8},

with V = ⨁_{i=1}^{8} ℝε_{i}. Let f(ε_{1}) = 23, f(ε_{2}) = 6, f(ε_{3}) = 5, … f(ε_{8}) = 0; we have an even number of odd f(ε_{i}), so f is integer on all roots. Then,

Δ_{+} = \{ε_{i} ± ε_{j}(i < j), \frac{1}{2} (ε_{1} ± ε_{2} ± ⋯ ± ε_{8})_{(even number of + signs)}\}_{i,j=1}^{8}

and

Π = \{ε_{2} − ε_{3}, ε_{3} − ε_{4}, … ε_{7} − ε_{8}, \frac{1}{2}(ε_{1} − ε_{2} − ⋯ − ε_{7} + ε_{8})\}.

In particular, f(α_{i}) = 1 for all i; θ = ε_{1} + ε_{2}, so α_{0} = −ε_{1} − ε_{2} and we get the diagrams E_{8} and Ẽ_{8} in Figure 1.1e.

Exercise 1. From Exercise 16.4, we have:

Δ_{E_{7}} = \{ε_{i} − ε_{j}(1 ≤ i ≠ j ≤ 8), \frac{1}{2}(±ε_{1} ± ε_{2} ± ⋯ ± ε_{8})_{4+signs}\}
Δ_{E_{6}} = \{ε_{i} − ε_{j}(1 ≤ i ≠ j ≤ 6), ±(ε_{7} − ε_{8}), \frac{1}{2}(±ε_{1} ± ε_{2} ± ⋯ ± ε_{6}_{3+signs}) ± \frac{1}{2}(ε_{7} − ε_{8}))\}.

Perform the same analysis as in the Examples for E_{7} and E_{6}. Show that their diagrams are as in Figure 1.1f,g.

For E_{7}, we will pick f(ε_{i}) so that all roots have non-zero integer values. For this, we need f(ε_{i}) to be distinct integers (hence ±ε_{i} ± ε_{j} are all integer and non-zero). Additionally, we want an even number of the f(ε_{i}) to be odd. That way, roots of the second type also have integer values. Take f(ε_{2}) = 7, f(ε_{3}) = 6, …, f(ε_{8}) = 1; then, α_{1} = ε_{2} − ε_{3}, α_{2} = ε_{3} − ε_{4}, …, α_{6} = ε_{7} − ε_{8} are all simple roots with value 1. We have (α_{i}, α_{j}) = −1 if |i − j| = 1 and 0 otherwise so far, and (α_{i}, α_{i}) ≡ 2. There is just one more simple root, and it must be of the second type. We need to choose f(ε_{1}) appropriately. In order for f to be integer-valued on all roots, f(ε_{1}) must be even. To avoid a zero-valued root, we will make it sufficiently large that

\frac{1}{2}(f(ε_{1}) − 7 − 6 − 5 − 4 + 3 + 2 + 1) > 0,

since that’s the closest we can get to zero if f(ε_{1}) is large. Simplifying, we have: f(ε_{1}) even, and f(ε_{1}) > 16, so we’ll take f(ε_{1}) = 18. Then f(α_{7}) = 1, α_{7} = \frac{1}{2}(ε_{1} − ε_{2} − ε_{3} − ε_{4} − ε_{5} + ε_{6} + ε_{7} + ε_{8}), and we have the last simple root. We have (α_{i}, α_{7}) = 0 for all 1 ≤ i ≠ 4 ≤ 6, with (α_{4}, α_{7}) = −1 and (α_{7}, α_{7}) = 2. Hence, all the connections are simple, and we have the right diagram. The highest root is clearly θ = ε_{1} − ε_{8} ↦^{f} 17, so α_{0} = ε_{8} − ε_{0}, which connects (simply) only with α_{7}, so we have the right extended diagram too.

For E_{6}, we choose the same f(ε_{2}), … f(ε_{8}). This gives us the following simple roots: α_{1} = ε_{2} − ε_{3}, …, α_{4} = ε_{5} − ε_{6}, α_{5} = ε_{7} − ε_{8}, with (α_{i}, α_{j}) = −δ_{ij} if 1 ≤ i, j ≤ 4; all (α_{i}, α_{i}) = 2, and α_{5} doesn’t connect with any of these roots. Now, to choose the right f(ε_{1}), we assume it’s large and take the smallest non-negative number we can get:

\frac{1}{2}(f(ε_{1}) − 7 − 6 − 5 + 4 + 3 − 2 + 1 > 0,

so f(ε_{1}) > 12. As before, we want it to be even, and so pick f(ε_{1}) = 14, which gives the last simple root f(α_{6}) = 1, α_{6} = \frac{1}{2}(ε_{1} −ε_{2} −ε_{3} −ε_{4} +ε_{5} +ε_{6} −ε_{7} +ε_{8}). α_{7} connects only with α_{3} and α_{5}; (α_{3}, α_{6}) = −1, (α_{5}, α_{6}) = −1 and (α_{6}, α_{6}) = 2. So, the connections are all simple, and we get the correct diagram. The highest root in this case is θ = ε_{1} − ε_{6} ↦^{f} 11, so α_{0} = ε_{6} − ε_{1}, which connects only with α_{4}, which, again, gives the right extended diagram.

Exercise 2. From Exercises 16.5 and 16.6, we have:

Δ_{F_{4}} = \{±ε_{i} ± ε_{j}, ±ε_{i}, \frac{1}{2}(±ε_{1} ± ε_{2} ± ε_{3} ± ε_{4})\}, 1 ≤ i ≠ j ≤ 4
Δ_{G_{2}} = {ε_{i} − ε_{j}, ±(ε_{i} + ε_{j} − 2ε_{k})}, 1 ≤ i, j, k ≤ 3, all distinct

Perform the same calculations for F_{4} and G_{2}, and show that they have the diagrams in Figure 1.1h,i.

Here, we proceed exactly as in the previous exercise. For F_{4}, we pick f(ε_{1}) = 8, f(ε_{2}) = 3, f(ε_{3}) = 2, f(ε_{4}) = 1, which gives us the simple roots α_{1} = ε_{2}−ε_{3}, α_{2} = ε_{3} − ε_{4}, α_{3} = ε_{4}, α_{4} = \frac{1}{2}(ε_{1} − ε_{2} − ε_{3} − ε_{4}), with simple connections between α_{1} and each of α_{2}, α_{3}, α_{4}. Additionally, there is a double-arrow connection from α_{2} to α_{3}, so we get the correct diagram. α_{0} = −ε_{1} − ε_{2} ↦^{f} −11 connects up only with α_{1}, which gets us the extended diagram.

For G_{2}, pick f(ε_{1}) = 4, f(ε_{2}) = 2, f(ε_{3}) = 1, which gives us simple roots α_{1} = ε_{2} − ε_{3} and α_{2} = ε_{1} + ε_{3} − 2ε_{2}, with (α_{1}, α_{1}) = 2,(α_{2}, α_{2}) = 6, and (α_{1}, α_{2}) = −3,

(a)  (b)  (c)  (d)

Figure 2.1. The four Dynkin diagram connection types, corresponding to the four types of 2 × 2 Cartan matrix minors.

r p q D̃_{4}  T_{p,q,r} (a)  (b)

Figure 2.2. Some diagrams for the simply laced case.

Figure 2.3. A loop with one double-connection.

which gives the desired connection. α_{0} = ε_{2} + ε_{3} − 2ε_{1} ↦^{f} −5, with a simple connection to α_{2}, as in the extended diagram.

2. Classification of Dynkin Diagrams

Theorem 2.1. The Dynkin diagrams of all indecomposable Cartan matrices are A_{r}, B_{r}, C_{r}, D_{r}, E_{8}, E_{7}, E_{6}, F_{4}, G_{2}.

Proof. We have to choose connected graphs with connections of the 4 types depicted in Figure 2.1, such that the matrix of any subgraph has a positive determinant. In particular, our graphs contain no extended Dynkin diagrams as induced subgraphs, since these have determinant 0. Part 1. Classify all “simply laced” Dynkin diagrams, i.e. using only 0- or 1-edge connections (which correspond to a symmetric A). Such a graph contains no cycles, since those are Ã_{r}. If there are no branching points, we get A_{r}. Next, such a graph contains at most 1 branching point, since otherwise it contains D̃_{r}. If there is a branching point, one has at most 3 branches, since D̃_{4} is the 4-star in Figure 2.2a. Thus, we are left with a graph of the form T_{p,q,r}(p ≥ q ≥ r ≥ 2) depicted in Figure 2.2b. But the graph cannot contain any of Ẽ_{6} = T_{3,3,3}, Ẽ_{7} = T_{4,4,2}, Ẽ_{8} = T_{6,3,2}, hence the only possibilities are D_{r}, E_{6}, E_{7}, E_{8}. Why? Suppose r = 3, then p ≥ q ≥ 3, so it contains T_{3,3,3}. Hence, r = 2. From Ẽ_{7}, we can’t have q ≥ 4 ⇒ q = 2, 3, and we get p ≤ 5 likewise.

Part 2. Classify all non-simply laced diagrams, i.e. those containing double- or triple-edge connections (corresponding to non-symmetric A). This can be done by a more complicated case analysis. However, we’d like an effective way of ruling out diagrams like the one in Figure 2.3. This can be done by computing many large determinants, but we would rather argue using the following:

Exercise 3. Prove the following two useful lemmas:

Lemma 2.2. Let

A = ( \begin{matrix}2 & \underline{−a 0 ⋯ 0} \\ −b ; 0 ; ⋮ ; 0 | & A_{n−1} \end{matrix} ) .

Then, det A = 2 det A_{n−1} − ab det A_{n−2}, where A_{n−2} is A with the first and second rows and columns deleted.

Proof. Denote by A\(i, j) the matrix A with the ith row and jth column removed. In particular, A_{n−2} = A\(1, 2)\(1, 1).

To prove this lemma, we expand the determinant by minors along the first row, getting det A = 2 det A_{n−1} + a det A\(1, 2); then, we expand det A\(1, 2) along its first column, which gives det A = 2 det A_{n−1} − ab det A_{n−2}, as desired. □

Lemma 2.3. Let

A = ( \begin{matrix}c_{1} & −a_{1} & 0 & ⋯ & 0 & −b_{n} \\ −b_{1} & c_{2} & −a_{2} & 0 & & 0 \\ 0 & −b_{2} & c_{3} & ... & ... & ⋮ \\ ⋮ & 0 & ... & ... & −a_{n−2} & 0 \\ 0 & & ... & −b_{n−2} & c_{n−1} & −a_{n−1} \\ −a_{n} & 0 & ⋯ & 0 & −b_{n−1} & c_{n} \end{matrix} ) .

Then, det(A − ϵE_{12}) = det A − ϵ(b_{1} det A_{n−2} + a_{2}a_{3} … a_{n}). In particular, if det A_{n−2} > 0, b_{1} > 0, a_{2} … a_{n} > 0, and ϵ > 0, then det A − ϵE_{12} < det A.

Proof. To prove this lemma, we expand the determinant of A − ϵE_{12} along the second column to get

(a_{1} + ϵ) det A\(1, 2) + c_{2} det A\(2, 2) + b_{2} det A\(3, 2) = det A + ϵ det A\(1, 2).

A\(1, 2) looks like this:

( \begin{matrix}−b_{1} & −a_{2} & 0 & ⋯ & 0 \\ 0 & c_{3} & ... & ... & ⋮ \\ ⋮ & ... & ... & −a_{n−2} & 0 \\ 0 & ... & −b_{n−2} & c_{n−1} & −a_{n−1} \\ −a_{n} & ⋯ & 0 & −b_{n−1} & c_{n} \end{matrix} ) .

If we expand its determinant along the first column, we get −b_{1} det A\(1, 2)\(1, 1)+ (−1)^{n−1}a_{n} det A\(1, 2)\(n−1, 1). The first part of the expression is simply −b_{1} det A_{n−2}.

det < 0 ; det < 0 ; det = 0 ; det = 0

Figure 2.4. Possible neighbors of G_{2} in a diagram.

As for the second part, the matrix we get is:

( \begin{matrix}−a_{2} & 0 & ⋯ & 0 \\ c_{3} & ... & ... & ⋮ \\ ... & ... & −a_{n−2} & 0 \\ ... & −b_{n−2} & c_{n−1} & −a_{n−1} \end{matrix} ) ,

which is lower triangular with a diagonal consisting of all a_{i}, 2 ≤ i ≤ n − 1, hence with determinant (−1)^{n−2}a_{2}a_{3} … a_{n−1}. Putting all this together, we get the desired

det A + ϵ(−b_{1} det A_{n−2} − a_{2}a_{3} … a_{n}).
□

Lemma 2.3 implies that in the non-simply laced case, there are no cycles as well. This is because det Ã_{r−1} = 0; so, any other cycle with subdiagrams of positive determinant must have determinant < 0 by the lemma. For example, the diagram in Figure 2.3 has A = Ã_{4} − E_{12}, so det A < 0.

Next, looking at the extended Dynkin diagrams we calculate that if the diagram contains G_{2} (Figure 1.1i) it must be G_{2}, by Lemma 2.2. Otherwise, it must contain one of the possibilities in Figure 2.4, none of which have suitable determinant: their Cartan matrices are of the form

A = ( \begin{matrix}2 & −a & 0 \\ −b & 2 & −1 \\ 0 & −3 & 2 \end{matrix} ) ⇒ A_{n−1} = ( \begin{matrix}2 & −1 \\ −3 & 2 \end{matrix} ) and A_{n−2} = ( 2 ) .

Hence, by Lemma 2.2, det(A) = 2 det(A_{n−1}) − ab det(A_{n−2}) = 2(1 − ab), which is zero or negative since a, b > 0. (The matrix of the second graph in the figure is actually A^{T}, but all the determinants are the same.)

It remains to look at the case when we have only simple or double connections. Looking at the extended Dynkin diagrams, C̃_{r} (in Figure 1.1c) cannot be a subdiagram. The variants with flipped arrow directions also don’t work. They are obtained from the extended Cartan matrix Ã of C̃_{r}by replacing some of Ã, Ã_{n−1}, Ã_{n−2} by their transposes, which doesn’t change any of the determinants in the calculation. Thus, by Lemma 2.2, their determinants are also 0.

Therefore only one double connection is possible. But, then we cannot have branching points, since B̃_{r} contains a double edge and a branching point. So, the only remaining case is a line with a left-right double edge, having p single edges to the left, and q single edges to the right. F̃_{4} has p = 2, q = 1; its transpose has p = 1, q = 2. Hence, if the diagram contains F̃_{4}, it must be F̃_{4}. Otherwise, either p = 0 or q = 0, and we have B_{r} or C_{r}.

□

We have now shown that any finite-dimensional simple Lie algebra yields one of a very restricted set of Dynkin diagrams (and hence Cartan matrices). The next step in the classification of semisimple Lie algebras will be to give a construction associating an abstract Cartan matrix to a Lie algebra, and hence to prove that these four classes plus five exceptional algebras are in fact the only finite-dimensional simple Lie algebras.