Abstract Nonsense

Crushing one theorem at a time

Relation Between the Number of Standard Young Tableaux on a Frame and the Number of Young Tableaux on the Frame’s Subordinate/Superordinate Frames


Point of Post: In this post we find a relation between the number of standard Young tableaux on a frame \mathcal{F} and the number of Young tableaux all the subordinate and superordinate frames to \mathcal{F}.

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Motivation

As was stated in our last post we can find a very interesting way to calculate the number, f_{\text{st}}\left(\mathcal{F}\right), of standard Young tableaux \mathcal{T} with \text{Frame}\left(\mathcal{T}\right)=\mathcal{F}. In this post we actually prove this claim. The intuitive idea is clear, by construction of Young tableaux we see that if \mathcal{T} is a Young Tableaux such that \text{Frame}\left(\mathcal{T}\right) is a n-frame then the number n must lie in a bottom right corner of \text{Frame}\left(\mathcal{T}\right) and then fixing n in that position b  we see that the possible Young tableaux are just the Young tableaux of \mathcal{F}-b and thus it makes sense then that f_{\text{st}}\left(\mathcal{F}\right) is some sort of sum of \text{f}_{\text{st}}\left(\mathcal{G}\right) where \mathcal{G} is taken over the subordinate frames to \mathcal{F}. The other theorem which has to do with finding f_{\text{st}}\left(\mathcal{F}\right) given the values f_{\text{st}}\left(\mathcal{H}\right) where \mathcal{H} is taken over the frames superordinate to \mathcal{F}.

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The Relation Between f_{\text{st}}\left(\mathcal{F}\right) and f_{\text{st}}\left(\mathcal{G}\right) for \mathcal{G}\triangleleft\mathcal{F}

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We jump right into the claim:

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Theorem: Let \mathcal{F} be a n-frame, then

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\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{G}\triangleleft \mathcal{F}}f_{\text{st}}\left(\mathcal{G}\right)

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Proof: Let \text{Tab}\left(\mathcal{F}\right) be the set of all Young tableaux with frame \mathcal{F}. Let b\in\mathcal{F} be the block with n\in b (by this I merely mean the block which contains the value n). Note that since the row b is increasing we must have that it is the last in its row since any block to its right would have to have a lower value than n. Similarly, b is the bottom of its columns for the same reason, and thus b\in\text{BCR}\left(\mathcal{F}\right). It clearly then follows that if we let for every b\in\text{BCR}\left(\mathcal{F}\right), \text{Tab}_b\left(\mathcal{F}\right) be the elements of \text{Tab}\left(\mathcal{F}\right) with n\in b then

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\displaystyle \text{Tab}\left(\mathcal{F}\right)=\bigsqcup_{b\in\text{BCR}\left(\mathcal{F}\right)}\text{Tab}_b\left(\mathcal{F}\right)\quad\mathbf{(1)}

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where the disjointness is obvious since elements from different \text{Tab}_b\left(\mathcal{F}\right)‘s have the value of n in different boxes. That said, it’s clear that every standard Young tableau in \text{Tab}_b\left(\mathcal{F}\right) is really just a standard Young tableau of \mathcal{F}-b (which we know is really a legitimate n-1-frame) with then just the extra condition that n\in b and moreover it’s clear that every standard Young tableau of \mathcal{F}-b can be extended to an element of \text{Tab}_b\left(\mathcal{F}\right) by taking the standard Young tableau, adding in the box b at it’s correct point and then filling it in with b. It clearly follows then that \#\text{Tab}_b\left(\mathcal{F}\right)=f_{\text{st}}\left(\mathcal{F}-b\right). Thus, by \mathbf{(1)} and our previous characterization of \text{Sub}\left(\mathcal{F}\right) we see that

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\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\#\text{Tab}\left(\mathcal{F}\right)=\sum_{b\in\text{BCR}\left(\mathcal{F}\right)}\#\text{Tab}_b\left(\mathcal{F}\right)=\sum_{b\in\text{BCR}\left(\mathcal{F}\right)}f_{\text{st}}\left(\mathcal{F}-b\right)=\sum_{\mathcal{G}\triangleleft\mathcal{F}}f_{\text{st}}\left(\mathcal{G}\right)

\blacksquare

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The Relation Between f_{\text{st}}\left(\mathcal{F}\right) and f_{\text{st}}\left(\mathcal{H}\right) for \mathcal{H}\triangleright\mathcal{F}

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Of course, as usual in algebra, there is a dual notion to the above relation which we dive right into

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Theorem: Let \mathcal{F} be a n-frame, then

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\displaystyle (n+1)f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)

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Proof: We proceed by induction on the size of the frames. For 1-frames this obvious since the unique 1-frame is \begin{array}{|c|}\cline{1-1}\hphantom{\times}\\ \cline{1-1}\end{array}. So, assume that the result is true for every n-frame and let \mathcal{F} be a n+1-frame. Then, using the previous theorem we note that

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\displaystyle \sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)=\sum_{H\triangleright\mathcal{F}}\sum_{\mathcal{K}\triangleleft\mathcal{H}}f_{\text{st}}\left(\mathcal{K}\right)

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In this second sum we are summing over subordinates of superordinates of \mathcal{F} and it’s clear that except the case when the superordinate is the addition of the new row at the bottom of \mathcal{F} every such subordinaet of a superordinate is really the superordinate of a subordinate of \mathcal{F}. Said differently

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\displaystyle \sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{H}\triangleright\mathcal{F}}\sum_{\mathcal{K}\triangleleft\mathcal{H}}f_{\text{st}}\left(\mathcal{K}\right)=\sum_{\mathcal{G}\triangleleft\mathcal{F}}\sum_{\mathcal{J}\triangleright\mathcal{G}}f_{\text{st}}\left(\mathcal{J}\right)+f_{\text{st}}\left(\mathcal{F}\right)

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But, since each \mathcal{J} is a n-frame (being subordinate to \mathcal{F}) we may apply the induction hypothesis and the first theorem to get

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\displaystyle \begin{aligned}\sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)& =\sum_{\mathcal{G}\triangleleft\mathcal{F}}\sum_{\mathcal{J}\triangleright\mathcal{G}}f_{\text{st}}\left(\mathcal{J}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=\sum_{\mathcal{G}\triangleleft\mathcal{F}}(n+1)f_{\text{st}}\left(\mathcal{G}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=(n+1)f_{\text{st}}\left(\mathcal{F}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=(n+2)f_{\text{st}}\left(\mathcal{F}\right)\end{aligned}

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and thus the induction is complete. \blacksquare

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

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May 11, 2011 - Posted by | Algebra, Algebraic Combinatorics, Representation Theory | , , , , , , , ,

3 Comments »

  1. […] make the double sum amenable to applying the dimension hypothesis. We’ll heavily use our two facts about the sum over subordinate and superordinate frames and the standard tableaux on them. So, […]

    Pingback by The Fundamental Result for Tableaux Combinatorics « Abstract Nonsense | May 12, 2011 | Reply

  2. […] proof is simple, we induct on the size of the frames (how many blocks it contains) and then use the relation between the number of standard Young tableaux on a frame and the number of standard Young tableaux […]

    Pingback by The Hook-length Formula « Abstract Nonsense | May 14, 2011 | Reply

  3. […] Recall that for a given -frame  we defined to be the set of all Young tableaux  with . What we first notice is that there is a natural -action on by permuting the numbers in the boxes. In particular, […]

    Pingback by Row and Column Stabilizer « Abstract Nonsense | May 20, 2011 | Reply


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