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Is there any general pattern for the roots of each of the classical lie algebras? So, can I tell all the roots of each of the $nth$ rank classical lie algebras $A_n, B_n, C_n, D_n$, as a linear combination of the simple roots. So I am looking for a pattern for the roots at each level given a classical lie algebra. So, if I am given $C_6$ for e.g., is there any general formula for the roots at say the $kth$ level in terms of $k$, e.g. $f(k,n)a_1+g(k,n)a_2+...$ are the roots, where $a_1,a_2,...$ are the simple roots.

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migrated from May 24 '12 at 7:31

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A summary:

For any root $\alpha$, the set of basis vectors $\alpha_i$ occuring in the expansion of $\alpha$ (i.e. the ones with non-zero coefficients) must form a connected set of nodes in the Dynkin diagram. For types $A,B,C,D$ only 0, 1 and 2 occur as coefficients. The coefficient 2 is forbidden in type $A$.

For type $A_n$ the sequences of coefficients are of the form $0^a1^{1+b}0^c$. Here $0^a$ denotes a string (possibly empty) of $a$ zeros, and $1^b$ a string of $b$ ones. Obviously $a,b,c$ are all non-negative integers subject to the constraint $a+b+c=n-1$.

For type $B_n$, in addition to the sequences that occurred in type $A_n$, also sequences of the form $0^a1^{1+b}2^{1+c}$ with $a+b+c=n-2$ are ok. Observe that there has to be at least a single 1 between the zeros and the 2s. The 2s are at the end of the lone short basic root (check your source: conceivably some may have the short basic root at the other end).

For type $C_n$, in addition to the sequences that occurred in type $A_n$, the sequences of the form $0^a1^b2^{1+c}1$ are ok. Again $a+b+c=n-2$. The string of 2s thus does not include, but is connected to, the lone long (rem. this was wrongly written as "short" in the earlier version) basic root.

For type $D_n$, any combination of 0s and 1s such that the 1s form a connected set in the Dynkin diagram is ok. In addition to that, sequences of the form $0^a1^{1+b}2^c\vphantom{(}^{\displaystyle{1}}_{\displaystyle{1}}$ with $a+b+c=n-3$ are ok. IOW the string of 2s must end at the vertex connected to three other vertices, and it must also have a "buffer" of at least a single 1 in each three directions separating the 2s from 0s.

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Thanks for the answer.+1 and accepted answer.! – user23238 May 25 '12 at 11:16
Thank you for very nice answer. However did your observation have any proof? I see a quick corollary that there always exists a number 1 in every sequence of coefficients. Do you have any quick explanation for my statement? – user Jun 22 '15 at 15:36

Yes, you can find all the roots of a Lie algebra as integer linear combinations of simple roots. That's because the generators corresponding to simple roots are enough to generate the whole group via commutators and their commutators etc. and the root associated with a commutator $[u,v]$ is simply the sum of the roots corresponding to the generators $u,v$ themselves.

In fact, with the usual definitions of simple roots – positive roots that can't be written as a sum of others – one may restrict the analysis to non-negative integer linear combinations of the simple roots when he reconstructs the whole root system.

For simply laced algebras $A_k$, $D_k$, $E_k$, all the (nonzero) roots are equally long. So given the metric on the Cartan subalgebra, the (nonzero) roots are simply all the integer linear combinations of the simple roots that have the same length as the simple roots (which are equally long, too). The algebras $B_k$, $C_k$, $F_4$, and $G_2$ are not simply laced so one must also include roots that are $\sqrt{2}$ times longer than a simple root (or $\sqrt{3}$ times longer in the $G_2$ case).

One shouldn't forget that $k$ (the rank) of the generators of a Lie algebra correspond to roots equal to zero – which are usually not counted among the roots at all and they are surely not considered simple roots.

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Sir, Please could you elaborate. I want all the roots of given algebra (e.g. $B_5$), without the use of a algorithmic procedure, but I am wondering if there is some pattern that I can use to directly tell the roots from the algebra. I have edited my question, to make it clearer. – user23238 May 23 '12 at 8:19
Sorry, your updated question got very confusing. I was answering your original question. The new question talks about "levels" etc. There aren't levels in ordinary Lie algebras. The roots are very simple objects and be sure I could easily enumerate all of them for all 7 classes of the Lie algebras, A,B,C,D,E,F,G. They're just the non-negative integral combinations of the simple roots that have the right length (either the same as simple roots or, in non-simply-laced groups, sqrt2 or sqrt3 times longer). ... If you have an algebra, the roots are just defined as the eigenvalues under the Cartan. – Luboš Motl May 23 '12 at 15:51
Are you sure that all the integer combination with the right roots are valid roots. For e.g. consider David's comment below..$α_3+α_4+α_5$ is a root but $α_3+α_4+α_6$ is not a root for $A_n$, though both are of same length (Are the not?). – user23238 May 24 '12 at 6:38
@ramanujan, those two are not of the same length. For type $A_n$ the answer is simple: The positive roots at level $k$ are sums of $k$ consecutive simple roots, i.e. sums of the form $$\sum_{j=i}^{i+k-1}\alpha_j$$ for some $i$ in the range $1\le i<n-k$. Here the level parameter $k$ ranges from $1$ to $n$. Is this what you are looking for? – Jyrki Lahtonen May 24 '12 at 8:07
@ramanujan: the simple roots are not always orthogonal (they are orthogonal, iff they are not connected in the Dynkin diagram). The root $\alpha_3+\alpha_4+\alpha_5$ is of the same length as all the roots. The vector $\alpha_3+\alpha_4+\alpha_6$ is $\sqrt{2}$ times as long, because the vectors $\alpha_6$ and $\alpha_3+\alpha_4$ are orthogonal. The latter vector is not a root of $A_n$, because $\{3,4,6\}$ is not a list of consequtive integers. The vector $\alpha_3+\alpha_4+\alpha_5+\alpha_6$ OTOH is a root. – Jyrki Lahtonen May 24 '12 at 13:12

I'll describe to you in the following a procedure to construct all the roots of a simple Lie algebra. This procedure is based on Slansky's review article. It is different from the method described in the Wikipedia page.

I think that it needs as little as possible input data.

Actually, the only inputs you need is the Dynkin diagram (Given in table 5 of Slansky), the highest weights of the adjoint representation (Given in table 8) and a few rules described in the sequel to make the construction.

  1. Construction of the Cartan matrix from the Dynkin diagram.The elements of the cartan matrix are the scalar product of the roots divided by the second root squared length.

1.1. Unconnected roots are orthogonal, the angle betwen two roots connected by one line is 120 degrees, 2 lines 135 degrees and three lines 150 degrees.

1.2. Only the relative lengths matter, it is conventional to take the squared length of the short roots (empty dots) as 1 and of the long roots (full dots) as 2 except for $G_2$ which is taken as 3.

1.3 Now that you have constructed the Cartan matrix, you can read the primitive roots which are given by the rows of the cartan matrix.

Remarks: These are the primitive root components in the weight basis in which the primitive weights have only one nonvanishing unit component).One must remenber that the weight space is not Eucledian. It has a metric that must be used in scalar products. The metric tensor can be constructed from the inverse of the Cartan matrix according to equation 4.11 in Slansky.

  1. Now that you have the primitive roots, one method to construct the whole root system is to find the weights of the adjoint representation. For this purpose, one can construct the weight diagram starting from the highest weight given in table 8 of Slansky using the method described on page 31, 32, the positive roots are just the positive weights of the adjoint representation.

2.1. The highest weight is picked from table 8.

2.2. From the highest weight and from any intermediate weight the n-th primitive root is subtracted a number of times equal to the n-th component of the weight if it is positive and not subtracted it it is zero or negative.

For example, in $A_3$ the roots $\alpha_1 = (2, -1, 0)$ and $\alpha_3 = (0, -1, 2)$ are subtracted each once from the highest weight $(1, 0, 1)$ in order to obtain the second level weights: $(-1 , 1, 1)$ and $(1, 1, -1)$.

Remark: This method allows you to construct the weight diagram but not to know the weight multiplicity, but in our case of the adjoint representation, we know that the multiplicity of all the weights is one because they are roots except for the zero weight whose multiplicity is equal to the rank of the Lie algebra.

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Sir, Thanks for the reply. But, what I want is not a systematic way for finding the roots. I am looking for a pattern for the roots at each level given a classical lie algebra. So, I am given $C_6$, is there any general formula for the roots at say the $kth$ level in terms of k, e.g. $f(k,n)a_1 + g(k,n)a_2 + ...$, where $a_1, a_2,... $ are the simple roots. – user23238 May 23 '12 at 12:53
I think that if one really tries to write formulas for the coefficients (e.g. $f(k,n)$), one will end up with very complex formulas which will look actually like computer programs. Please try to think about the simplest case of $A_n$ where the roots are sums of continuous strings of primitive roots (e.g. $\alpha_3+\alpha_4+\alpha_5$ is a root but $\alpha_3+\alpha_4+\alpha_6$ is not). Now, you can see that to write the coefficient of say $\alpha_1$ at a certain level is very complicated. (cont.) – David Bar Moshe May 23 '12 at 13:23
(cont.) This is the reason that representation theory uses algorithms (very frequently combinatorial ones) to perform the constructions rather than closed formulas. – David Bar Moshe May 23 '12 at 13:23

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