Mixing
(Un)boundedness of real-valued homomorphism on some lattice.
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Let $mathbbZ^k$ denote the finitely-generated free abelian group on $k$ generators and consider an injective homomorphism $$chi colon (mathbbZ^k,+) to (mathbbR,+).$$ Now let $(a_n)$ be a sequence of pairwise distinct elements in $mathbbZ^k$. My question is:
Does there exist a subsequence of $(a_n)$ such that $vert chi(a_n) vert$ tends to $+infty$?
The answer for $k=1$ is yes as then $chi$ corresponds to a non-zero multiplication and one can extract a subsequence of $(a_n) subseteq mathbbZ$ either going towards $+ infty$ of $-infty$.
For $k>1$ this seems more subtle and I'm not sure anymore whether the answer to the question is affermative. Maybe the following bit might be helpful:
Observe that $chi$ can be viewed as a vector $(chi_1, dots, chi_k) in mathbbR^k$ such that $$chi(b^1,dots b^k)=langle (chi_1, dots,chi_k),(b^1,dots,b^k) rangle=sum_i=1^k chi_i cdot b^i.$$ Injectivity of $chi$ then implies that the $(chi_i)_i=1^k subseteq mathbbR$ are linearly independent over $mathbbQ$.
linear-algebra abstract-algebra combinatorics group-homomorphism
asked Jul 18 at 11:59
noctusraid
8381626
add a comment |Â
up vote
2
down vote
favorite
Let $mathbbZ^k$ denote the finitely-generated free abelian group on $k$ generators and consider an injective homomorphism $$chi colon (mathbbZ^k,+) to (mathbbR,+).$$ Now let $(a_n)$ be a sequence of pairwise distinct elements in $mathbbZ^k$. My question is:
Does there exist a subsequence of $(a_n)$ such that $vert chi(a_n) vert$ tends to $+infty$?
The answer for $k=1$ is yes as then $chi$ corresponds to a non-zero multiplication and one can extract a subsequence of $(a_n) subseteq mathbbZ$ either going towards $+ infty$ of $-infty$.
For $k>1$ this seems more subtle and I'm not sure anymore whether the answer to the question is affermative. Maybe the following bit might be helpful:
Observe that $chi$ can be viewed as a vector $(chi_1, dots, chi_k) in mathbbR^k$ such that $$chi(b^1,dots b^k)=langle (chi_1, dots,chi_k),(b^1,dots,b^k) rangle=sum_i=1^k chi_i cdot b^i.$$ Injectivity of $chi$ then implies that the $(chi_i)_i=1^k subseteq mathbbR$ are linearly independent over $mathbbQ$.
linear-algebra abstract-algebra combinatorics group-homomorphism
asked Jul 18 at 11:59
noctusraid
8381626
Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $mathbbZ^k$ denote the finitely-generated free abelian group on $k$ generators and consider an injective homomorphism $$chi colon (mathbbZ^k,+) to (mathbbR,+).$$ Now let $(a_n)$ be a sequence of pairwise distinct elements in $mathbbZ^k$. My question is:
Does there exist a subsequence of $(a_n)$ such that $vert chi(a_n) vert$ tends to $+infty$?
The answer for $k=1$ is yes as then $chi$ corresponds to a non-zero multiplication and one can extract a subsequence of $(a_n) subseteq mathbbZ$ either going towards $+ infty$ of $-infty$.
For $k>1$ this seems more subtle and I'm not sure anymore whether the answer to the question is affermative. Maybe the following bit might be helpful:
Observe that $chi$ can be viewed as a vector $(chi_1, dots, chi_k) in mathbbR^k$ such that $$chi(b^1,dots b^k)=langle (chi_1, dots,chi_k),(b^1,dots,b^k) rangle=sum_i=1^k chi_i cdot b^i.$$ Injectivity of $chi$ then implies that the $(chi_i)_i=1^k subseteq mathbbR$ are linearly independent over $mathbbQ$.
linear-algebra abstract-algebra combinatorics group-homomorphism
asked Jul 18 at 11:59
noctusraid
8381626
Let $mathbbZ^k$ denote the finitely-generated free abelian group on $k$ generators and consider an injective homomorphism $$chi colon (mathbbZ^k,+) to (mathbbR,+).$$ Now let $(a_n)$ be a sequence of pairwise distinct elements in $mathbbZ^k$. My question is:
Does there exist a subsequence of $(a_n)$ such that $vert chi(a_n) vert$ tends to $+infty$?
The answer for $k=1$ is yes as then $chi$ corresponds to a non-zero multiplication and one can extract a subsequence of $(a_n) subseteq mathbbZ$ either going towards $+ infty$ of $-infty$.
For $k>1$ this seems more subtle and I'm not sure anymore whether the answer to the question is affermative. Maybe the following bit might be helpful:
Observe that $chi$ can be viewed as a vector $(chi_1, dots, chi_k) in mathbbR^k$ such that $$chi(b^1,dots b^k)=langle (chi_1, dots,chi_k),(b^1,dots,b^k) rangle=sum_i=1^k chi_i cdot b^i.$$ Injectivity of $chi$ then implies that the $(chi_i)_i=1^k subseteq mathbbR$ are linearly independent over $mathbbQ$.
linear-algebra abstract-algebra combinatorics group-homomorphism
asked Jul 18 at 11:59
noctusraid
8381626
asked Jul 18 at 11:59
noctusraid
8381626
asked Jul 18 at 11:59
noctusraid
8381626
asked Jul 18 at 11:59
noctusraid
8381626
8381626
Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24
add a comment |Â
Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24
Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24
add a comment |Â
1 Answer
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oldest
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1
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I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$eta = sum_k=0^infty 2^-2^k$$
Since its binary expansion does not repeat, $eta$ must be irrational. for all $nge 0$, let $$q_n = 2^2^n,quad p_n = lfloor q_neta rfloor$$
Then $$0 le q_neta - p_n = 2^2^nsum_k=n+1^infty 2^-2^k < 2^2^nsum_l=2^n+1^infty 2^-l = 2^2^n2^1-2^n+1 = 2^1-2^n to 0$$
as $n to infty$.
Therefore the map $chi : Bbb Z^2 to Bbb R : (n, m) mapsto n + meta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) times (q_n)$ such that $chi(a_n) to 0$ as $n to infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^2^n$.
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$eta = sum_k=0^infty 2^-2^k$$
Since its binary expansion does not repeat, $eta$ must be irrational. for all $nge 0$, let $$q_n = 2^2^n,quad p_n = lfloor q_neta rfloor$$
Then $$0 le q_neta - p_n = 2^2^nsum_k=n+1^infty 2^-2^k < 2^2^nsum_l=2^n+1^infty 2^-l = 2^2^n2^1-2^n+1 = 2^1-2^n to 0$$
as $n to infty$.
Therefore the map $chi : Bbb Z^2 to Bbb R : (n, m) mapsto n + meta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) times (q_n)$ such that $chi(a_n) to 0$ as $n to infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^2^n$.
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
add a comment |Â
up vote
1
down vote
accepted
I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$eta = sum_k=0^infty 2^-2^k$$
Since its binary expansion does not repeat, $eta$ must be irrational. for all $nge 0$, let $$q_n = 2^2^n,quad p_n = lfloor q_neta rfloor$$
Then $$0 le q_neta - p_n = 2^2^nsum_k=n+1^infty 2^-2^k < 2^2^nsum_l=2^n+1^infty 2^-l = 2^2^n2^1-2^n+1 = 2^1-2^n to 0$$
as $n to infty$.
Therefore the map $chi : Bbb Z^2 to Bbb R : (n, m) mapsto n + meta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) times (q_n)$ such that $chi(a_n) to 0$ as $n to infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^2^n$.
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$eta = sum_k=0^infty 2^-2^k$$
Since its binary expansion does not repeat, $eta$ must be irrational. for all $nge 0$, let $$q_n = 2^2^n,quad p_n = lfloor q_neta rfloor$$
Then $$0 le q_neta - p_n = 2^2^nsum_k=n+1^infty 2^-2^k < 2^2^nsum_l=2^n+1^infty 2^-l = 2^2^n2^1-2^n+1 = 2^1-2^n to 0$$
as $n to infty$.
Therefore the map $chi : Bbb Z^2 to Bbb R : (n, m) mapsto n + meta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) times (q_n)$ such that $chi(a_n) to 0$ as $n to infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^2^n$.
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$eta = sum_k=0^infty 2^-2^k$$
Since its binary expansion does not repeat, $eta$ must be irrational. for all $nge 0$, let $$q_n = 2^2^n,quad p_n = lfloor q_neta rfloor$$
Then $$0 le q_neta - p_n = 2^2^nsum_k=n+1^infty 2^-2^k < 2^2^nsum_l=2^n+1^infty 2^-l = 2^2^n2^1-2^n+1 = 2^1-2^n to 0$$
as $n to infty$.
Therefore the map $chi : Bbb Z^2 to Bbb R : (n, m) mapsto n + meta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) times (q_n)$ such that $chi(a_n) to 0$ as $n to infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^2^n$.
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
answered Jul 19 at 23:13
Paul Sinclair
18.5k21439
18.5k21439
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
add a comment |Â
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
very nice, thanks alot!
– noctusraid
Jul 19 at 23:17
add a comment |Â
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Can your figure out how to prove that there exist integer sequences $(p_n), (q_n)$ such that $p_n - q_npi to 0$ as $n to infty$?
– Paul Sinclair
Jul 18 at 23:26
@PaulSinclair yeah, there exists such a sequence as $mathbbQ subseteq mathbbR$ is dense! Got it, thanks!
– noctusraid
Jul 18 at 23:42
It's a little stronger than just density of $Bbb Q$. Instead of $pi$, it would be easier to prove for an irrational number such as $$sum_n 2^-2^n$$.
– Paul Sinclair
Jul 19 at 12:23
@PaulSinclair Don't we just need that if $fracp_nq_n to pi$ (with positive integers $p_n,q_n$), then $p_n$ and $q_n$ tend to $+ infty$?
– noctusraid
Jul 19 at 17:34
No, because $q_n$ can grow faster than $fracp_nq_n - pi$ shrinks, in which case $p_n - q_npi$ would also grow. By carefully choosing the two sequences, you probably can avoid this when using $pi$. But it is practically obvious that you can avoid it for the other irrational number.
– Paul Sinclair
Jul 19 at 19:24