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Let $S_k$ be the space of weight 12 cusp forms of $\Gamma_0(p)$, ($p$ prime), then Sage tells that $\dim S_k^{\text{new}}=\dim S_k-2$. Thus the old forms spans a 2-dimensional subspace. One of the obvious eigenbase vectors is $\Delta(z)=q\prod_{n=1}^\infty(1-q^n)^{24}$. What does the other eigenbase vector look like?

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    $\begingroup$ The other is $\Delta(pz)$ $\endgroup$ – Peter Humphries Feb 5 '16 at 19:32
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    $\begingroup$ Is $\Delta(pz)$ an eigenform of Hecke operator? $\endgroup$ – Dianbin Bao Feb 5 '16 at 19:34
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    $\begingroup$ Yes, with all the same eigenvalues as $\Delta(z)$ (perhaps except for the Hecke operator $T_p$). $\endgroup$ – John Binder Feb 5 '16 at 20:26
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    $\begingroup$ It's maybe also worth noting that you don't need Sage to see this; it follows form the very definition of $S^{\text{new}}$: this is the space orthogonal to the space of oldforms, and the oldforms are those coming from cusp forms of $\Gamma_0(n)$ with $n\mid p$, $n\neq p$. In this case, the only such $n$ is $1$, the unique eigenform on $SL_2$ is $\Delta(z)$, and the only forms that come from it at level $p$ are $\Delta(z)$ and $\Delta(pz)$. $\endgroup$ – John Binder Feb 5 '16 at 20:29
  • $\begingroup$ @JohnBinder: I suggest that you turn your comments into an answer, so that this question can be closed. $\endgroup$ – GH from MO Feb 6 '16 at 0:24
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Per suggestion of GH from MO in the comments, I'm turning my comment into an answer.

In general, the space $S_{k}(\Gamma_0(N))$ decomposes as $S_{k}^{\text{new}}(\Gamma_0(N)) \oplus S_{k}^{\text{old}}(\Gamma_0(N))$. Here the space of oldforms is spanned by the forms $f(dz)$, where $f$ is a newform of weight $k$ and level $\Gamma_0(N')$, with $N' < N$ and $(N'\cdot d) \mid N$. The space $S^{\text{new}}$ is the orthogonal complement of $S^{\text{old}}$ under the Petersson inner product. Moreover, for a newform $f$ at level $N'$, the forms $f(dz)$ and $f(d'z)$ have the same Hecke eigenvalues (away from primes dividing $N$).

Now we apply this to the situation at hand, where $k = 12$ and $N = p$ is prime. There is only one form of weight $12$ and level $1$, which is $\Delta$ (up to scalar multiple). Since $p$ is prime, all oldforms must arise from forms of level $1$, so the space of oldforms has dimension $2$: it is spanned by $\Delta(z)$ and $\Delta(pz)$, and both have the same Hecke eigenvalues away form $p$.

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