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In the introduction of this paper, the authors say that: Let $M \in \mathbb{Z}_{>0}$. If $f$ is a normalised newform for $\Gamma_0(M)$ then we define $$\Lambda(s,f)=(2\pi)^{-s}\Gamma(s) M^{s/2}L(s,f)$$

It satisfies a functional equation $$ \Lambda(s,f)=\epsilon \Lambda (k-s,f)$$ where $\epsilon=\pm 1$. The order of vanishing of $L(s,f)$ at $s=k/2$ is even if $\epsilon=1$, and is odd if $\epsilon=-1$.

I do not understand the bold statement. In the paper, the authors say it is "clearly" but I find it is not indeed trivial.

Desunkid
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1 Answers1

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First note that the order of vanishing of $\Lambda$ is the same as that of $L$ at $s=k/2$.

Now assume $\Lambda(k/2,f)=C \neq 0$ (ie order of vanishing is $0$). Then the functional equation gives $C=\epsilon C$ hence $\epsilon = 1$ as $C \neq 0$.

If the order of vanishing is $1$, then differentiate the functional equation to get $0 \neq \Lambda'(s,f) = \epsilon (-\Lambda'(k-s,f))$ hence substituting $s=k/2$ this time means $\epsilon = -1$.

More generally, if the order of vanishing is $n$, differentiate $n$ times to get $\Lambda^{(n)}(s,f)=\epsilon (-1)^n \Lambda^{(n)}(k-s,f)$ and hence $\epsilon=(-1)^n$.

Matt B
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