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Chris Wuthrich
Chris Wuthrich
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vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvLet $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$ but also no reason to believe they do not exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Let $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$ but also no reason to believe they do not exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.

Post Deleted by Chris Wuthrich
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Chris Wuthrich
Chris Wuthrich
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Let $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$, but also no reason why they don't exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Let $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$, but also no reason why they don't exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

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Chris Wuthrich
Chris Wuthrich
  • 8.9k
  • 1
  • 36
  • 47

Let $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$, but also no reason why they don't exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.