## Notations and Recollections

For a partition , one takes its Young diagram comprising of boxes. A *filling* is given by a function for some positive integer *m*. When *m*=*d*, we will require the filling to be bijective, i.e. *T* contains {1,…,*d*} and each element occurs exactly once.

If and is a filling, then is obtained by replacing each *i* in the filling with *w*(*i*). For a filling *T*, the corresponding row (*resp*. column) tabloid is denoted by {*T*} (*resp*. [*T*]).

Recall from an earlier discussion that we can express the -irrep as a quotient of from the surjection:

Here is any fixed bijective filling .

Concretely, a **C**-basis for is given by column tabloids [T] and the quotient is given by relations: where *T’* runs through all column tabloids obtained from *T* as follows:

- fix columns
*j*<*j’*and a set*B*of*k*boxes in column*j’*of*T*; then*T’*is obtained by switching*B*with a set of*k*boxes in column*j*of*T*, while preserving the order. E.g.

## For Representations of GL_{n}

_{n}

From the previous article we have , where is the quotient of the space of column tabloids described above. We let be the set of all functions , i.e. the set of all fillings of λ with elements of *V*. We define the map:

for any bijective filling This is independent of the *T* we pick; indeed if we replace *T* by for , the resulting RHS would be:

where the first equality holds since the outer tensor product is over and the second equality follows from our definition . Hence is well-defined. It satisfies the following three properties.

Property C1. is multilinear in each component V.

In other words, if we fix and consider as a function on *V* in component *s* of , then the resulting map is **C**-linear. E.g. if , then:

This is clear.

Property C2. Suppose are identical except and , where are in the same column. Then

**Proof**

Let be the transposition swapping *s* and *t*. Then by alternating property of the column tabloid and . Thus:

♦

Finally, we have:

Property C3. Let Fix two columns in the Young diagram for λ, and a set B of k boxes in column j’. As A runs through all sets of k boxes in column j, let be obtained by swapping entries in A with entries in B while preserving the order. Then:

E.g. for any we have:

**Proof**

Fix a bijective filling Then:

where swaps the entries in *A* with those in *B* while preserving the order (note that ). But the sum of all such vanishes in Hence ♦

## Universality

Definition. Let V, W be complex vector spaces. A map is said to beλ-alternatingif properties C1, C2 and C3 hold.The

universal λ-alternatingspace (or theSchur module) for V is a pair where

- is a complex vector space;
- is a λ-alternating map,
satisfying the following universal property: for any λ-alternating map to a complex vector space W, there is a unique linear map such that

*F*(*V*) is not hard to construct: the universal space which satisfies C1 and C2 is the alternating space:

So the desired *F*(*V*) is obtained by taking the quotient of this space with all relations obtained by swapping a fixed set *B* of coordinates in with a set *A* of coordinates in , and letting *A* vary over all |*A*| = |*B*|. E.g. the relation corresponding to our above example for C3 is:

over all

By universality, the λ-alternating map thus induces a linear:

You can probably guess what’s coming next.

Main Theorem. The above is an isomorphism.

## Proof of Main Theorem

First observe that is surjective by the explicit construction of *F*(*V*) so it remains to show injectivity via dim(LHS) ≤ dim(RHS).

Now , and we saw earlier that its dimension is the number of SSYT with shape λ and entries in [*n*].

On the other hand, let be the standard basis of If *T* is any *filling* with shape λ and entries in [*n*], we let be the element of *F*(*V*) obtained by replacing each *i* in *T* by ; then running through the map

Claim. The set of generates , where T runs through all SSYT with shape λ and entries in [n].

**Proof**

Note that the set of , as *T* runs through all *fillings* with shape λ and entries in [*n*], generates *F*(*V*).

Let us order the set of all fillings of *T* as follows: *T’* > *T* if, in the rightmost column *j* where *T’* and *T* differ, at the lowest in which , we have .

This gives a total ordering on the set of fillings. We claim that if *T* is a filling which is not an SSYT, then is a linear combination of for *S* > *T*.

- If two entries in a column of
*T*are equal, then by definition.

- If a column
*j*and row*i*of*T*satisfy , assume*j*is the rightmost column for which this happens, and in that column,*i*is as large as possible. Swapping entries and of*T*gives us*T’*>*T*and

- Now suppose all the columns are strictly ascending. Assume we have , where
*j*is the largest for which this happens, and , for . Swapping the topmost*i*entries of column*j*+1, with various*i*entries of column*j*, all the resulting fillings are strictly greater than*T*. Hence , where each*S*>*T*.

Thus, if *T* is not an SSYT we can replace with a linear combination of where *S* > *T*. Since there are finitely many fillings *T* (with entries in [*n*]), this process must eventually terminate so each can be written as a linear sum of for SSYT *S*. ♦

Thus ≤ number of SSYT with shape λ and entries in [*n*], and the proof for the main theorem is complete. From our proof, we have also obtained:

Lemma. The set of forms a basis for F(V), where T runs through the set of all SSYT with shape λ and entries in [n].