Lower bound for sum of Hecke eigenvalues
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
add a comment |
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
edited Nov 20 at 14:14
asked Nov 12 at 13:17
Nodt Greenish
30513
30513
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
add a comment |
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
edited Nov 22 at 8:40
answered Nov 20 at 14:12
Nodt Greenish
30513
30513
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995298%2flower-bound-for-sum-of-hecke-eigenvalues%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown