Two important comments on the scalar product

1. Again, please note the mistake I made in class and on the lecture notes. I am about to correct the notes and reupload them:

https://hopfcombinatorics.wordpress.com/2012/04/26/latex-se_lambda-1n-h_lambda-and-a-mistake-in-lecture-3/

2. The inner product on Sym has the very nice property that it is respected by the antipode; that is, <Sf,Sg> = <f,g>. You may find this useful in the last question of the Short Homework. **If** one of you posts a proof of this fact here, then everyone will be allowed to use it in the homework (and will be very thankful to that person for their effort).

 

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7 comments

  1. Here’s my stab at it: Clearly, the inner product induces a norm. Let’s define it as follows: \sqrt{\langle f, g \rangle} = ||f - g||. Then we have, for any given f, g \in Sym

    ||Sf - Sg||^2 = ||S(f - g)||^2 = ||S(\sum (f_i - g_i) e_i)||^2

    \rightarrow = ||\sum (-1)^i f_i h_i - \sum (-1)^i g_i h_i||^2 = ||f - g||^2.

    Hence, ||Sf - Sg|| = ||f - g||, which implies the claim.

  2. could someone direct me to the definition of the inner product on symmetric functions.

  3. p.106 of the notes.

  4. duquec, I don’t think that’s how you want to define a norm. Also, is our inner product positive definite? (It is, but why?)

    Miriam, please note that I haven’t fixed the notes on the website yet – the definition there is wrong at the moment. You want to use the definition here:
    https://hopfcombinatorics.wordpress.com/2012/04/26/latex-se_lambda-1n-h_lambda-and-a-mistake-in-lecture-3/
    where we define the inner product between the m and the h, and then extend bilinearly.

  5. alyssa palfreyman

    Here’s my try at a proof:

    It is enough to show \langle f,g\rangle = \langle S(f), S(g)\rangle for basis elements. In particular, I show

    \langle e_\lambda, h_\mu\rangle = \langle S(e_\lambda), S(h_\mu)\rangle.

    In his book Enumerative Combinatorics, Stanley provides a way to write the basis \{e_\lambda\} in terms of \{m_\mu\} as

    e_\lambda = \sum_{\mu\vdash n} M_{\lambda\mu}m_\mu

    where \lambda \vdash n (proposition 7.4.1). For a description of what M_{\lambda\mu} is, I direct you to Stanley’s book – it has something to do with the number of matrices satisfying certain conditions. The important property of these numbers is that
    M_{\lambda\mu} = M_{\mu\lambda}

    (corollary 7.4.2). We know from class that

    \langle m_\lambda,h_\mu\rangle = \langle h_\mu,m_\lambda\rangle = \begin{cases} 1 \quad\text{ if } \mu = \lambda\\ 0 \quad\text{otherwise}\end{cases}

    Suppose \lambda is a partition of k and \mu is a partition of l. Then the left hand side is

    \langle e_\lambda, h_\mu\rangle = \langle \sum_{\nu\vdash k}M_{\lambda\nu}m_\nu, h_\mu\rangle = \begin{cases} M_{\lambda\nu} = M_{\lambda\mu}\quad\text{if } \nu = \mu \\ 0 \qquad\qquad\text{otherwise}\end{cases}

    and the right hand side is

    \langle S(e_\lambda), S(h_\mu)\rangle = \langle (-1)^kh_\lambda, (-1)^le_\mu\rangle = (-1) ^{k+l} \langle h_\lambda, \sum_{\tau\vdash l}M_{\mu\tau}m_\tau\rangle = \begin{cases} M_{\mu\tau} = M_{\mu\lambda}\quad \text{if }\lambda = \tau \\ 0 \qquad\qquad\text{otherwise}\end{cases}

    Notice that if \lambda = \tau (or if \nu = \mu) then it must be that k=l (otherwise they would be partitions of different numbers and not equal). And the right hand side is equal to the left hand side.

  6. Thank you, Alyssa, especially from your classmates. 🙂
    You should all feel free to use it in your solution to the homework.

    I have one objection to your argument: I did not prove that this inner product is symmetric ( = ), which you are using in the proof. Can anyone fill in the details to that?

  7. ninalyssa

    Thanks, Alyssa! Maybe we can work out Federico’s objection tomorrow in the office.

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