Young Tableaux Subs

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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 ofS, and also representations ofsu(n).

(Relations between this use of and n will be explained later.) Thei-th row has i boxes, and

these numbers are constrained so that

1 2 3 . . . ,

i=1

i= . (1.1)

It is clear from the above that the sequence {i}

i=1

constitutes a partition of the integer. The

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

if a particular value of a i 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

1 for the number of cycles of length 1, 2 for the number of cycles of length 2, etc., up to

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:

j =

k=j

k , k =k k+1 ,

k=1

kk =n =

j=1

j . (1.3)


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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 ofthe 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 repre-

sentation ofS, 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. Thehook-lengthof 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 4 3 2 15 3 2 11

4 3 12 1

5 2 1

21

4 2

3 11

5 13

21

543

21

1 4 5 6 5 4 1

. (1.4)

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 ofS corresponding to that Young diagram. As an example

we consider the partitions ofS3, 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 ofS 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

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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 re-

maining, namely the one corresponding to the Young diagram , which is a 2-dimensional

representation. The two standard tableaux associated with this are1 23 and

1 32 .

II. For Simple Lie Algebras

Recall that a simple Lie algebra has a Cartan subalgebra, of rank r, say, and an associated

root space. This root space has a vector-space basis in terms of its simple roots, {i}ri=1, and

an alternate basis in terms of the reciprocal basis vectors, {k}rk=1, which satisfy

2k, j

J, J=kj . (2.1)

We describe the irreducible representations of such an algebra via its highest weight vec-

tor, , which is a linear combination of the reciprocal basis vectors with integer coefficients,

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

m11 +m2

2 + . . . +mrr = [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

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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

namedj 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 whenmr = 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:

number of boxes for the irrep isr

j=1

j mj . (2.3)

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

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

tensors, the Kronecker delta k

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

or alternating, (weighted) tensor abc...n or abc...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

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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):

dimension of = n p

m n divided by

3 22 1

= 112n

2(n2 1) with p n+ 1

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:

n p q divided by 3 2 1 = 16n(n+ 1)(n+ 2) ,

nmk

divided by321

= 16n(n 1)(n 2),

n pm

divided by 3 1

1 = 1

3n(n2 1),

q n+ 2 ,

k n 2,

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

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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 ob-

jects 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 representa-

tions 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

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

components, {a}na=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 subspacesas 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.

As a vector space this direct product, V V, has a basis {ea eb}na,b=1. Invariant subspacesare the symmetric parts and the skew-symmetric partson the indices a and b, as seen in

ea eb = 12 (ea eb+ eb ea) + 12 (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,

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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 ofV, each of which contains the fundamental representation of

su(n), and use all the irreps ofS 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 ofsu(n) a Young diagram with an arbitrary number of columns but at most n1

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 ofn indices, and imagining inserting numbers into the

various Young diagrams, associated with S, and therefore having a total ofboxes in such

a way that we first take the symmetric part in each row separately, andafter having done

thattake 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 ofsu(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 VV

into its invariant subspaces, we can portray all this as follows, using the irreps ofS3:

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

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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

sym 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 ofsu(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 ofS3, , 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

skew 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 sub-

space 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 symbolsince 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

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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 S3is . 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 spacecreated by the Young symmetrizing and anti-symmetrizing rules

acting as directed by those two distinct standard tableaux:

1 23

(3) ea eb ec+ eb ea ec ec eb ea ec ea eb ,1 3

2 (4) ea eb ec+ ec eb ea eb ea ec eb ec ea .

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:

ea eb ec= 16{sym+ skew+ 2((3)+ (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 Youngdiagram 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

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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 ofn nmatrices, 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 ofn

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 .

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b. Next one takes the first box containing anaand appends it to the first Young diagram in

all possible ways that follow the rules for creation of a Young diagram:

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 ana and do the same thing with it, exceptthat we

are not allowed to put twoas together in the same column. The reason for this is that the

original as 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 a a 1 aa 1 aa

1aa

b .

Now we notice that the last diagram in the large braces does indeed have two as in the

same column; therefore, it must be deletedor, 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 as and or bs 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:

1 a a

1 aa

b .

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

1 a a b 1 a a

b

1 a ba

1 aa b

1 aab

.

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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 as so far encountered,

the number of bs so far encountered, the number of cs so far encountered, etc. If, at

any step in this procedure, the number ofbs encountered so far exceeds the number

of as encountered, or if the number of cs encountered exceeds the number of bs, 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 columnfor su(3)and the usual labeling for the irreps in the column

below that:

=

3 8 = 15 + 6 + 3

[1, 0] [1, 1] = [2, 1] [0, 2] [1, 0]

. (2.7)

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

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recall that the Cartan matrix for su(n) is given by the matrix (n 1) (n 1) matrix:

A=

2 1 0 0 0 . . . 0 0 01 2 1 0 0 . . . 0 0 00 1 2 1 0 . . . 0 0 0

... 0 1 2 1 . . . 0 0 0

...... 0 1 2

. . . 0 0 0

. . . . . . . . . . . . . . . . . . . . . . . . . . .

0 0 0 0 0 . . . 2 1 00 0 0 0 0 . . . 1 2 10 0 0 0 0 . . . 0 1 2

(2.8)

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:

A1 = 1

n

n 1 n 2 n 3 n 4 . . . 3 2 1n 2 2(n 2) 2(n 3) 2(n 4) . . . 6 4 2n 3 2(n 3) 3(n 3) 3(n 4) . . . 9 6 3n 4 2(n 4) 3(n 4) 4(n 4) . . . 12 8 4

......

......

. . . ...

......

3 6 9 12 . . . 3(n 3) 2(n 3) n 32 4 6 8 . . . 2(n 3) 2(n 2) n 21 2 3 4 . . . n 3 n 2 n 1

. (2.9)

Therefore we may immediately write

1 = 1n

{(n 1)1+ (n 2)2+ (n 3)3+ . . . + 2n2+ n1} ,

2 = 1n

{(n 2)1+ 2(n 2)2+ 2(n 3)3+ . . . + 4n2+ 2n1} ,

. . .

n1 = 1n

{1+ 22+ 33+ . . . + (n 2)n2+ (n 1)n1} .

(2.10)

We also consider the representation with highest weight 1 =1 = [1, 0, 0, . . . , 0], and the

system of weights that lies below it. The next lowest weight is obviously

1 1 =1 1=

1n

{1+ (n 2)2+ (n 3)3+ . . . + 2n2+ n1} . (2.11a)

The next one after that is

1 1 2 =1 1 2 =

1n

{1 22+ (n 3)3+ . . . + 2n2+ 1} . (2.11b)

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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 to 1 1, we can see that the sum is just 2, 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.

Again if we look above we may add 1 to 1 1 and also 1 1 2; doing this we acquire

3, 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 1, 2, etc., i.e., one where there are j 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 1, of

the representation [0, 1, 0, . . . , 0] is 2, with [0, 0, 1, 0, . . . , 0] is 3, etc. As well we have seen

above that 2 = 1 1, and also that 3 = 1 1 2 =

2 2. Investigation of the

elements ofA1 will assure you that this approach continues all the way down. Note that the

fundamental representation has exactlyn weights, with the last one being 1 n1

k=1k. Now

return to the Young diagram being considered. To find the highest weight associated with it

we first fill the entire first row with 1, acceptable because that row is symmetrized. We next

fill the entire second row with the next-lower weight, also called the second-highest weight,which is 1 1. The third highest weight,

1 1 2 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

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

box contains 1 1 2. This continues until the last box, the j-th one, which contains

1 1 . . . j1. The highest weight for the entire diagram is the contributions from all

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

1 (1 1) (1 1 2) . . . (

1 1 . . . j1)

= j1 (j 1)1 (j 2)2 . . . 2j2 j1 =j ,

(2.12)

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

also only true ifj 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 j , we can easily see

that the highest weight for the entire Young diagram is given by

Young=m11 +m2

2 + . . . +mn1n1 =

n1j=1

mjj (2.13)

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.

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