Some Notes on Young Tableaux As Useful for Irreps of Su(N)

Some Notes on Young Tableaux as useful for irreps of su(n)

I. For the Symmetric Groups: A Young diagram is a collection of rows of boxes, stacked vertically on top of each other,

left-justiﬁed. They are used to discuss representations of S`, and also representations of su(n).

(Relations between this use of ` and n will be explained later.) The i-th row has λi boxes, and these numbers are constrained so that X` λ1 ≥ λ2 ≥ λ3 ≥ ... ≥ λ` , λi = ` . (1.1) i=1 ` It is clear from the above that the sequence {λi}i=1 constitutes a partition of the integer `. The

usual notation for a partition is simply this sequence of values of the λi’s, except that, often,

if a particular value of a λi is repeated one simply symbolizes this by putting an appropriate power on that value. For instance, the partitions of 5 are as follows, and correspond to the Young diagrams given below each partition: (5) (4, 1) (3, 2) (3, 12) (22, 1) (2, 13) (15)

. (1.2)

It is also common to present elements of the symmetric group, S`, in terms of cycles, of lengths 1, 2, 3, 4, . . . , ` − 1, `, where one usually does not actually write out those of length 1, since a cycle of length 1 eﬀects no change on its entries. Therefore one also wants to know how many cycles of which length are involved in a particular group element. This is noted by the symbols

ν1 for the number of cycles of length 1, ν2 for the number of cycles of length 2, etc., up to ν` for the number of cycles of length `, where this last quantity obviously can only have the value of 0 or 1. These two sets of integers are related as follows: X` λj = νk , νk = λk − λk+1 , k=j X` X` kνk = n = λj . (1.3) k=1 j=1 Clearly each set is sufficient to determine the other set; however, they have different useful purposes. Among these is the fact that for a given value of ` the set of all Young diagrams, or of all partitions, displays all possible ways of defining ways of dividing up products of ` copies of some fixed number of objects.

Each partition of ` determines a Young diagram which determines an irreducible representation of S`, and all irreducible representations are determined in this way. The dimension of an irreducible representation associated with a given Young diagram is determined via the product of the hook-lengths of all its elements. The hook-length of a given box in a given Young diagram is the sum of all boxes in its row to its right plus itself plus all boxes in its own column below it. If the product of all these hook-lengths is divided into `!, one obtains the dimension of that irreducible representation. Examples for the 7 Young diagrams given above for S5 are as below, with the hook-length boxes in the second row and the dimensions of the irreducible representations in the third row: (5) (4, 1) (3, 2) (3, 12) (22, 1) (2, 13) (15) 5 5 1 4 5 2 1 4 2 3 3 5 3 2 1 4 3 1 2 3 1 2 2 . (1.4) 5 4 3 2 1 1 2 1 1 1 1 1

1 4 5 6 5 4 1 A Young tableau is a Young diagram with numbers inserted into it, from 1 to `. A standard Young tableau is one where the numbers inserted all increase from left to right, along any row, and also from up to down, along any column. The set of all possible standard Young tableaux

form a basis for the representation of S` corresponding to that Young diagram. As an example

we consider the partitions of S3, which has only 3 classes, and therefore 3 irreps. These include the two (non-faithful) 1-dimensional ones: the trivial one where all elements are represented by +1, and the alternating one where elements are represented by their sign, either +1 or −1.

The sign of an element of S` is determined by viewing it in its cycle format: those that have an odd number of cycles with an even number of elements within them are negative

2 and are represented by −1 in this (second) 1-dimensional representation, while the others are positive and are represented by +1. This is always exactly half of the entire group, and the

positive ones form a subgroup, referred to by A`, which is a simple, ﬁnite group. It should be noted that these two representations (clearly) occur for every value of `.

Returning, however, to the value ` = 3, there is only one irreducible representation remaining, namely the one corresponding to the Young diagram , which is a 2-dimensional 1 2 1 3 representation. The two standard tableaux associated with this are 3 and 2 .

II. For Simple Lie Algebras Recall that a simple Lie algebra has a Cartan subalgebra, of rank r, say, and an associated

r root space. This root space has a vector-space basis in terms of its simple roots, {αi}i=1, and k r an alternate basis in terms of the reciprocal basis vectors, {η }k=1, which satisfy

k hη , αji k 2 = δj . (2.1) hαJ , αJ i

We describe the irreducible representations of such an algebra via its highest weight vector, Λ, which is a linear combination of the reciprocal basis vectors with integer coeﬃcients,

r {mj}j=1, often referred to as Dynkin integers:

1 2 r Λ ≡ m1η + m2η + ... + mrη −→= [m1, m2, . . . , mr] , (2.2)

where the notation on the right-hand side is our usual scheme for naming the distinct irreps, or irreducible representations, via the sequence of these integers. For every non-negative integer

choice of the values for each of the mj’s, for j running from 1 to r, we obtain a distinct irreducible representation of the algebra.

There are several questions that one often wishes to ask concerning the properties of irreducible representations, two of the most common being the dimension of that representation and the decomposition of the direct product of two or more of such representations. It turns out that a very useful tool to answer these questions is provided by associating with each irrep

3 a Young diagram. We make this assignment by using each mj as the number of columns

of length j in the associated Young diagram, i.e., we identify these mj’s with the quantities named νj in our discussion of the creation of the Young diagrams to study the representations of the symmetric group. I note here that a derivation of the form of this assignment is given toward the end of these notes. As examples, the fundamental representation [1, 0, 0,..., 0] is associated with a Young diagram with only a single box, . Then the representation

[0, 1, 0,..., 0], with m2 = 1, has only a single column with two boxes in it, stacked vertically, while [0, 2, 0,..., 0] would have two such columns next to each other. Finally the representation [0, 0,..., 0, 1] would have a Young diagram with a single column with r boxes stacked up since it corresponds to all mj = 0 except for j = r when mr = 1. Note that an algebra of rank r can have a representation with arbitrarily large positive integer values for any of the mj’s, so that the associated Young diagram may have as many boxes as one wants. More precisely it will have the following number of boxes as determined by the values of the highest weight vector:

Xr number of boxes for the irrep Λ is j mj . (2.3) j=1

However, the maximum value for j is r, so that the maximum length of any column is only r boxes.

Although in principle all the further usages for the description of irreps via Young diagrams could be done for any simple Lie algebra, it is especially convenient to use it for the algebras

Ar which have real, compact form su(n), where n = r + 1, or sl(n,R) as one of the non- compact real forms. The reason for doing is that these algebras have only two invariant

k tensors, the Kronecker delta δ` , which accomplishes traces but only when one has both upper (contravariant) and lower (covariant) indices on tensors, and the n-dimensional Levi-Civita,

abc...n or alternating, (weighted) tensor ² or ²abc...n, which accomplishes skew-symmetrization and can also be used to lower, or raise, indices but does not maintain constant their number. It is of course invariant because all the group representation matrices are required to have determinant +1. The other sets of simple Lie groups also contain “metrics” which are used

4 to raise and lower indices, which then complicates the use of Young diagrams. Therefore we will only consider the problem for su(n), which will suﬃciently well describe the process without going into those additional complications generated by the additional invariant tensors possessed by other sequences of Lie algebras.

An immediate use of this association of su(n), where r = n − 1, with Young diagrams is a quick method to calculate the dimension of any irreducible representation. This method is a generalization of the method used for using Young diagrams to calculate the dimension of irreps of the symmetric group, which uses the hook-length of a given diagram, as described in the neighborhood surrounding Eq. (1.4). The dimension is given by the ratio of two products of integers. The denominator is in fact the hook length of the associated diagram; however, the numerator is to be calculated according to the following scheme. Insert the integer n into the uppermost box in the diagram, and then insert other integers into the remaining boxes so that they increase by 1 as one goes along a given row, left to right, and they decrease by 1 as one goes along a given column, from up to down. The following is an example for the representation [0, 2, 0,..., 0] for su(n):   p ≡ n + 1 n p 3 2 dimension of = divided by = 1 n2(n2 − 1) with , m n 2 1 12  m ≡ n − 1

(where I use the symbology m = n − 1 and p = n + 1 only because they “ﬁt” better into the Young diagram boxes.) Another few examples may also be useful:

1 n p q divided by 3 2 1 = 6 n(n + 1)(n + 2) ,  n 3 1  q ≡ n + 2 , m divided by 2 = 6 n(n − 1)(n − 2) , k 1  k ≡ n − 2 , n p 3 1 divided by = 1 n(n2 − 1) , m 1 3

We now want to consider direct products of representations. Generally we would consider any particular representation as a group of operators acting on vectors, or states, that describe some physical objects. When we want to consider systems of such objects they will be acted

5 upon by systems of the representation in question. Mathematically this requires the creation of direct products of those representations. Depending on how many copies of the original objects are to be included in the system we will need direct products of that many copies of the original representation. In principle this can be done for copies of any original representation, and most everything below would apply. However, since for su(n) all irreducible representations can be obtained from direct products of the fundamental n-dimensional representation of labeled by [1, 0, 0,..., 0] we will only consider decompositions of ` copies of this particular representation, which will explain the method well enough. In more detail, we note that the representation acts via automorphisms of an n-dimensional vector space V , the carrier space

n for the representation. After having chosen a basis {ea}a=1 for that space, the group representations may be presented as n × n matrices which act on the vectors, described via their

a n components, {φ }a=1, which, presumably, relate to some set of states of some physical objects. If we then want to consider systems of these objects, we need to consider direct products of the vector spaces, and therefore of the representations. Such direct products are much more useful when they are subdivided into their invariant subspaces—as this usually provides the system with sets of useful parameters that describe them, corresponding to the various irreps that occur in the products. Mathematically this corresponds to ﬁnding invariant subspaces within the direct product spaces.

We begin with the simplest case, namely the direct product of two copies of the spaces.

n As a vector space this direct product, V ⊗ V , has a basis {ea ⊗ eb}a,b=1. Invariant subspaces are the symmetric parts and the skew-symmetric parts—on the indices a and b, as seen in

1 1 ea ⊗ eb = 2 (ea ⊗ eb + eb ⊗ ea) + 2 (ea ⊗ eb − eb ⊗ ea) . (2.4)

This decomposition corresponds to the two diﬀerent Young diagrams associated with S2, namely and However, when we consider products of more than two copies of V there are other invariant subspaces that occur, in addition to those that are completely symmetric or completely anti-symmetric. The determination of these other sorts of invariant subspaces,

6 on a set of indices, is a principal purpose of the symmetric group, S`, which determines all the distinct, irreducible ways to combine a set of ` objects in invariant ways. Therefore we immediately invoke it for this purpose. Rigorous proofs that this is what is needed can be found in the literature, but we presume here that it makes good sense on the basis of our understanding of the irreps of the symmetric group and do not go further in that direction but simply use it toward our purposes.

We now consider ` copies of V , each of which contains the fundamental representation of

su(n), and use all the irreps of S` to determine the invariant subspaces of this direct product vector space. It should be speciﬁcally noted that this is quite a distinct use of the Young

diagrams in their relationship to the Lie algebra being considered. Earlier we associated with an

arbitrary irrep of su(n) a Young diagram with an arbitrary number of columns but at most n−1

rows, as a diﬀerent and very useful way to label the particular irreducible representation, with

the n − 1 Dynkin integers being used to deﬁne the number of boxes in each column. However,

now we are considering ` copies of a set of n indices, and imagining inserting numbers into the various Young diagrams, associated with S`, and therefore having a total of ` boxes in such a way that we ﬁrst take the symmetric part in each row separately, and—after having done

that—take the skew-symmetric part in each column separately. The ﬁnal result will (almost

always) no longer be symmetric but will be skew-symmetric. The number of distinct standard

tableaux for a given diagram will then correlate with the dimension of the associated irreducible

representation of su(n) contained in this direct product of ` copies of the fundamental irrep.

Looking, as a simple example, at ` = 3, i.e., at the decomposition of the vector space V ⊗V ⊗V

into its invariant subspaces, we can portray all this as follows, using the irreps of S3:

a. The subspace characterized by , which corresponds to the totally symmetric part

of the 3 vector indices. Since the irrep is totally symmetric it can contain 3 copies of

the highest weight of the original, so that it is the irrep denoted by [3, 0,..., 0]. This is

consistent with the fact that the Young diagram for the original irrep was just , so that

7 this is simply 3 copies of that irrep, placed together in a single row since they are being considered symmetrically. The corresponding basis for this invariant subspace is then

θsym ≡ ea ⊗ eb ⊗ ec + ea ⊗ ec ⊗ eb + eb ⊗ ea ⊗ ec + eb ⊗ ec ⊗ ea + ec ⊗ eb ⊗ ea + ec ⊗ ea ⊗ eb .

Do note that if we had begun with a diﬀerent original representation of su(n), say the one associated with , then this resulting symmetric subspace would have had the Young diagram . Therefore it is an “accident” of the fact that we begin with just [1, 0,..., 0] that the Young diagram for the resulting irrep is the same as the diagram for the division of the copies into their invariant parts. However, it is a convenient accident from the point of view of calculation; it is just that it is easy to confuse them philosophically as well. b. For the next invariant subspace we choose this irrep of S3, , which will give us the totally skew-symmetric part of the original direct product. Since we have begun with just ,

this resultant subspace is also denoted by . Since it has only the one column, of length 3, it is also denoted by [0, 0, 1, 0,..., 0], with basis for its invariant subspace as

θskew ≡ ea ⊗ eb ⊗ ec − ea ⊗ ec ⊗ eb − eb ⊗ ea ⊗ ec + eb ⊗ ec ⊗ ea − ec ⊗ eb ⊗ ea + ec ⊗ ea ⊗ eb ,

again consistent with the idea that the columns of the Young diagram are skew-symmetrized. One should however note that if we were dealing with su(2) there would be no such subspace since there can only be one box in the Young diagram associated with an algebra with rank 1. As well, were we considering su(3), a diagram with 3 skew symmetric entries would have to be proportional to the Levi-Civita symbol—since the indices a, b, c, etc., only take on the values from 1 to 3 = n. As the Levi-Civita symbol is an invariant for this group this particular diagram would correspond simply to an invariant, i.e., to the 1- dimensional representation, usually referred to as scalars. Therefore, for su(3) this Young diagram is simply a substitute symbol for the 1-dimensional, non-faithful representation. Only for n ≥ 4 is the above statement literally true; of course that is consistent with the

8 fact that the presentation form we are using for the Dynkin coeﬃcients only has n − 1 entries, so there would be no “place” for an entry involving columns of length 3 until n ≥ 4. c. The last remaining irrep for S3 is . It has two standard tableaux, i.e., as a representation

for S3 it is two dimensional; therefore, it generates distinct two sets of basis vectors for the direct product space—created by the Young symmetrizing and anti-symmetrizing rules acting as directed by those two distinct standard tableaux:

1 2 → θ ≡ e ⊗ e ⊗ e + e ⊗ e ⊗ e − e ⊗ e ⊗ e − e ⊗ e ⊗ e , 3 (3) a b c b a c c b a c a b 1 3 → θ ≡ e ⊗ e ⊗ e + e ⊗ e ⊗ e − e ⊗ e ⊗ e − e ⊗ e ⊗ e . 2 (4) a b c c b a b a c b c a

One ﬁnds that this is indeed a decomposition of the original tensor product basis set for the entire space via the following equality in the space V ⊗ V ⊗ V :

1 ea ⊗ eb ⊗ ec = 6 {θsym + θskew + 2(θ(3) + θ(4))} . (2.5)

It is worth returning to the discussion of the statement that the diagram with just one column, just above in our decomposition of the triple product of only indicated the 1- dimensional representation if we were considering su(3). This is symptomatic of a more general problem with these associations of Young diagrams with irreps when we are multiplying them together. We noted originally that the associated diagram for su(n) would never have a column with more than n − 1 = r boxes since the number of reciprocal basis vectors is only n − 1. Nonetheless, when considering products of representations it could easily occur that a column with more than n − 1 boxes will appear. Remembering that the various columns of a Young diagram are skew-symmetrized before we present that Young diagram as being associated with any given irrep of a Lie algebra, if we begin with vector spaces where the indices run from 1 to, say, w, then we cannot skew-symmetrize more than w of them at once. Therefore any resultant representation that contains more than w boxes in a single column does not really exist; it is simply an artefact of the mode of construction, and must be thrown out. In the

9 event that we are considering just products of the fundamental representation denoted by , we have w = n, and no diagrams that contain columns with more than n boxes must simply be deleted. Then, if a column is created that contains exactly n boxes, then it contains the complete skew-symmetrization of n values for the indices; such a construct is always simply proportional to the Levi-Civita symbol, the standard skew-symmetric (weighted) tensor for n indices. However, this is of course the symbol that creates determinants of n × n matrices, and all of the matrices involved with the Lie group SU(n) are required to have determinant +1, so that this symbol is left invariant. Therefore the representation associated with a column of n boxes is simply a (non-faithful) representation in an invariant 1-dimensional subspace, i.e., a scalar that does not transform under the group but stays the same.

We can move on to what is perhaps the most important use of the Young diagrams as labels for the irreducible representations: they may be used to calculate the coeﬃcients in a

Clebsch-Gordon series for direct products of irreps. We now explain the rules by which this calculation is made. A derivation of these rules may be found, for instance, in an appendix to the text by Sternberg noted in the list of books interesting for this course, although the derivation could be called “arcane.” As well his statement of the rule about how to delete various summands in the product that violate some rules is diﬃcult to follow. The rules are also very well explained in volume 2 of the text by Cornwell, who has a better explanation for the rule mentioned just above, except of course that he forgets to append a diﬀerent important rule involving skew-symmetrization, which I will note below when I describe it. a. First one writes down the two Young diagrams for the two irreps under consideration,

such as in the following example. In the second one needs to insert a series of the letter a

in the ﬁrst row, the letter b in the second row, the letter c in the third row, etc, in order

to keep track of them once they are included in the various resultant diagrams:

a a ⊗ . b

10 b. Next one takes the ﬁrst box containing an a and appends it to the ﬁrst Young diagram in all possible ways that follow the rules for creation of a Young diagram: ½ ¾ 1 a 1 a ⊕ a ⊗ b ,

where I have inserted a symbol 1 into the Young diagram of the ﬁrst box, only because I

have so far been unable to cause the TEXmacros to type it otherwise. It is not necessary for the explication of the rules, but, oh, well. c. Next we take the next box containing an a and do the same thing with it, except that we are not allowed to put two a’s together in the same column. The reason for this is that the original a’s were symmetrized and the ﬁnal columns are to be skew-symmetrized. This is also the particular rule that Cornwell forgot to mention as clearly as he might have.   1   1 a 1 a a 1 a a ⊕ ⊕ ⊕ ⊗ b .  a a a 

Now we notice that the last diagram in the large braces does indeed have two a’s in the same column; therefore, it must be deleted—or, better, never have been entered because we knew that it would have to be deleted. As well we notice that the second and third entries are identical; therefore, we only count it once instead of twice. Were the two identical diagrams to have had a different pattern of a’s and or b’s within them they would have counted as different, and both would have been kept. Therefore, having deleted unacceptable diagrams the current result is just the following, where the step after this is to bring in the final box, containing a b: ½ ¾ 1 a 1 a a ⊕ ⊗ b . a

The result of bringing in that box with a b is the following:

1 a a 1 a b 1 a 1 a 1 a a b ⊕ ⊕ ⊕ ⊕ . b a a b a b

11 d. We are now ready for the next set of rules. The important one is as follows. One makes

a count, proceeding from right to left along the 1st row, and then in the same direction

along the 2nd row, etc., counting, at each box, the number of a’s so far encountered,

the number of b’s so far encountered, the number of c’s so far encountered, etc. If, at

any step in this procedure, the number of b’s encountered so far exceeds the number

of a’s encountered, or if the number of c’s encountered exceeds the number of b’s, etc.,

etc., then that diagram must be deleted from the sum, to avoid double-counting. Looking

back at the result just above, this rule tells us that the ﬁrst and third of the diagrams

above must be deleted. This leaves with the remaining three as the direct product desired,

noting, as well, that none of these remaining three are similar so that all should be kept.

We therefore now present the ﬁnal result, removing all the symbols inside since they were

simply a device to obtain the correct results:

⊗ = ⊕ ⊕ . (2.6)

As a quick, simple check on this we apply it to su(3). In that case we must ﬁrst take

the last diagram, which begins with a column with 3 boxes, and delete that column since

it would be invariant under SU(3). I then repeat that result below, with the dimensions

given in the next column—for su(3)—and the usual labeling for the irreps in the column

below that:

⊗ = ⊕ ⊕

3 × 8 = 15 + 6 + 3 . (2.7)

[1, 0] ⊗ [1, 1] = [2, 1] ⊕ [0, 2] ⊕ [1, 0]

Let us now think a bit more about the weights of a given representation of su(n). We

12 recall that the Cartan matrix for su(n) is given by the matrix (n − 1) × (n − 1) matrix:   2 −1 0 0 0 ... 0 0 0    −1 2 −1 0 0 ... 0 0 0     0 −1 2 −1 0 ... 0 0 0   .   . 0 −1 2 −1 ... 0 0 0     . . ..  A =  . . 0 −1 2 . 0 0 0  (2.8)    ..   ......     0 0 0 0 0 ... 2 −1 0   0 0 0 0 0 ... −1 2 −1  0 0 0 0 0 ... 0 −1 2 It is therefore possible to explicitly calculate the inverse of this matrix, which can then be used to determine those linear combinations of the simple roots that determine the reciprocal roots, and thereby the highest weights of the irreducible representations:  n − 1 n − 2 n − 3 n − 4 ... 3 2 1   n − 2 2(n − 2) 2(n − 3) 2(n − 4) ... 6 4 2     n − 3 2(n − 3) 3(n − 3) 3(n − 4) ... 9 6 3     n − 4 2(n − 4) 3(n − 4) 4(n − 4) ... 12 8 4  −1 1   A =  ......  . (2.9) n  ......     3 6 9 12 ... 3(n − 3) 2(n − 3) n − 3    2 4 6 8 ... 2(n − 3) 2(n − 2) n − 2 1 2 3 4 . . . n − 3 n − 2 n − 1 Therefore we may immediately write

1 1 η = n {(n − 1)α1 + (n − 2)α2 + (n − 3)α3 + ... + 2αn−2 + αn−1} ,

2 1 η = n {(n − 2)α1 + 2(n − 2)α2 + 2(n − 3)α3 + ... + 4αn−2 + 2αn−1} , (2.10) ...

n−1 1 η = n {α1 + 2α2 + 3α3 + ... + (n − 2)αn−2 + (n − 1)αn−1} . We also consider the representation with highest weight Λ1 = η1 −→= [1, 0, 0,..., 0], and the system of weights that lies below it. The next lowest weight is obviously

1 1 1 Λ − α1 = η − α1 = n {−α1 + (n − 2)α2 + (n − 3)α3 + ... + 2αn−2 + αn−1} . (2.11a)

The next one after that is

1 1 1 Λ − α1 − α2 = η − α1 − α2 = n {−α1 − 2α2 + (n − 3)α3 + ... + 2αn−2 + α1} . (2.11b)

13 Now, thinking back to the fact that the skew-symmetric portion of the product of [1, 0,..., 0]

with itself is [0, 1, 0,..., 0] we know that the highest weight for that skew-symmetric portion

would be the highest weight of [1, 0,..., 0] plus the second-highest weight. Therefore, if we add

1 1 2 η to η − α1, we can see that the sum is just η , showing that all this can be accomplished by yet other means. To continue further in this direction the skew-symmetric portion of the

triple product of the fundamental representation would have highest weight equal to the sum of

the original highest weight plus the second-highest weight plus, yet, the third-highest weight.

1 1 1 Again if we look above we may add η to η − α1 and also η − α1 − α2; doing this we acquire η3, which corresponds to [0, 0, 1, 0,..., 0], as indeed it should since it is associated with the

Young diagram .

Now let us take this investigation just a little further. We suppose that we are given an

irreducible representation of su(n) which is associated with a Young diagram according to

the partition λ1, λ2, etc., i.e., one where there are λj boxes in the j-th row, with j running no higher than n − 1. We now want to determine the highest weight associated with this representation. We know that the highest weight of the fundamental representation is η1, of

the representation [0, 1, 0,..., 0] is η2, with [0, 0, 1, 0,..., 0] is η3, etc. As well we have seen

2 1 3 1 2 above that η = η − α1, and also that η = η − α1 − α2 = η − α2. Investigation of the elements of A−1 will assure you that this approach continues all the way down. Note that the

1 Pn−1 fundamental representation has exactly n weights, with the last one being η − k=1 αk. Now return to the Young diagram being considered. To find the highest weight associated with it we first fill the entire first row with µ1, acceptable because that row is symmetrized. We next

ﬁll the entire second row with the next-lower weight, also called the second-highest weight,

1 1 which is µ − α1. The third highest weight, µ − α1 − α2 enters into all the boxes in the third row. This continues until we run out of boxes. Now let us consider what are the contents

1 1 of a column of length j. Its top box contains η ; the second box contains η − α1; the third

1 box contains η − α1 − α2. This continues until the last box, the j-th one, which contains

1 η − α1 − ... − αj−1. The highest weight for the entire diagram is the contributions from all

14 the boxes; however, we see that the contribution of the j-th column is just the sum above:

1 1 1 1 η − (η − α1)− (η − α1 − α2) − ... − (η − α1 − ... − αj−1) (2.12) 1 j = jη − (j − 1)α1 − (j − 2)α2 − ... − 2αj−2 − αj−1 = η ,

where the last equality is seen explicitly by studying the inverse matrix above. As well it is

also only true if j ≤ n − 1. This is ﬁne, of course, since no columns are allowed to have more

than n − 1 boxes. Since every column with j boxes contributes a term ηj, we can easily see that the highest weight for the entire Young diagram is given by

nX−1 1 2 n−1 j ΛYoung = m1η + m2η + ... + mn−1η = mjη (2.13) j=1

The above counting process can now be seen as an explicit justification of the assignment of a specific Young diagram to a specific irreducible representation that was made at the beginning of this section.