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I am reading a paper Bispectral and ($\mathfrak{gl}_n, \mathfrak{gl}_m$) dualities. I have some questions about some computations with Wronskian and dimensions of some vector spaces.

On page 9 (line -2), Section 3.5, why the space has dimension $\bar{n}+N$?

On page 10 (line 4), why $F_{k}(z)$ has dimension $k$?

Many thanks.

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The first question (re pg. 9, ln. -2) is about counting the basis elements $\{x^je^{\lambda_i x} \;|\; 1 \leq i \leq N, 0 \leq j \leq n_i\}$. Now the claim in the paper is only true when all $\lambda_i$ are distinct, and a quick perusal of the paper tells me that is true. Ok, so let's count elements! When $i=1$, there are the elements $x^0e^{\lambda_1 x}, x^1e^{\lambda_1 x}, \ldots, x^{n_1}e^{\lambda_1 x}$, for a total of $1 + n_1$. Similarly, for each $i$, there are $1 + n_i$ basis elements. Then the total will be $N\cdot 1 + n_1 + n_2 + \cdots n_N$, which is $\overline{n} + N$ according to their definition of $\overline{n}$.

As for $F_k(z)$, I'm not sure what's going on, since the paper states "$F_i(z)$ has dimension $k$". Clearly there's a typo, and they might have intended "$F_k(z)$ has dimension $k$". Even assuming that edit, I'd be a little suspicious of the statement. What is the dimension of $F_1(z)$? (In other words, can you verify $dim(F_1(z)) = 1$?) Are there properties of the $\lambda_i$'s that shed light on the nature of $F_k(z)$?

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I think that is a typo. Is the product of any subset of $e^{\lambda_1x}, \ldots, x^{n_1}e^{\lambda_1x}$ in $X$? –  LJR Jul 27 '11 at 17:50
    
@user9791: Not as I understand it. As a vector space, only sums of scalar multiples of the $x^je^{\lambda_i x}$ would lie in $X$. Think of $X$ as the solution space of some (very large) ODE. –  Shaun Ault Jul 27 '11 at 20:21

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