0
$\begingroup$

Use Ferrers diagrams to show bijectively that the number of self-conjugate partitions of $n$ is the same as the number of partitions of $n$ whose parts are odd and distinct. An example of the latter would be this partition of $12: {7+5}$. But, ${7+3+2}$ doesn’t work – they are distinct but not all odd.

I need help constructing the bijection and with a layman's definition of what self-conjugate means. Any help is appreciated!

$\endgroup$
  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – Brian Mar 12 at 1:36
  • 1
    $\begingroup$ Is this what you are looking for en.wikipedia.org/wiki/… ? $\endgroup$ – Donald Splutterwit Mar 12 at 1:40
0
$\begingroup$

Hint: look at the hooks whose central square is on the diagonal of a self conjugate partition.

$\endgroup$
0
$\begingroup$

I can think of two ways to do this:

  • Natural bijection between Ferrers diagrams
  • Naturally biject each kind of diagram with a specific kind of multiset of integers.

You can find a natural bijection between the odd partitions and the self-conjugate partitions. This is easiest to see if we number the dots in the Ferrers diagrams. In the case of the self-conjugate partitions, we'll label each dot with the number of steps you need to take to reach the left edge of the diagram. For the partition with odd, distinct parts we'll count the number of steps needed to reach the top.

You can convert a partition on the left to a partition on the right by collecting all of the dots whose distance from the top-left boundary is $x$ and placing them on the $x$th row of the right. This process is invertible, so we have a bijection.

Alternatively, we can biject both the left and right set of diagrams with multisets of non-negative integers where each element in the multiset has positive, odd multiplicity and smaller elements have strictly greater multiplicity than larger elements.

n    self-conjugate diagram    multiset              odd & distinct parts diagram
1    0                         {0}                   0

2    n/a                       n/a                   n/a

3    0 0                       {0,0,0}               0
     0

4    0 0                       {0,0,0,1}             0 0 0
     0 1                                             1

5    0 0 0                     {0,0,0,0,0}           0 0 0 0 0
     0
     0

6    0 0 0                    {0,0,0,0,0,1}          0 0 0 0 0
     0 1                                             1
     0

9    0 0 0                    {0,0,0,0,0,1,1,1,2}    0 0 0 0 0 
     0 1 1                                           1 1 1
     0 1 2                                           2
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.