Dual of a vector space coming from a lattice

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Let $M$ be a lattice i.e., $M$ is an abelian group and $Mcongmathbb Z^n$ for some $n$. Let $N:=hom_mathbb Z(M,mathbb Z)$. Let $M_mathbb R$ denotes the $mathbb R$-vector space $Motimes_mathbb Zmathbb R$. Then clearly $M_mathbb Rcongmathbb R^n$.

I need to show that $N_mathbb R:=Notimes_mathbb Zmathbb R$ is the dual of $M_mathbb R$, i.e., I need to show that $N_mathbb R:=hom_mathbb Z(M,mathbb Z)otimes_mathbb Zmathbb Rconghom_mathbb R(M_mathbb R,mathbb R)$.

Any hint will be helpful.

Thank you.

vector-spaces commutative-algebra abelian-groups

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asked Jul 26 at 13:05

2015

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up vote
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Let $M$ be a lattice i.e., $M$ is an abelian group and $Mcongmathbb Z^n$ for some $n$. Let $N:=hom_mathbb Z(M,mathbb Z)$. Let $M_mathbb R$ denotes the $mathbb R$-vector space $Motimes_mathbb Zmathbb R$. Then clearly $M_mathbb Rcongmathbb R^n$.

I need to show that $N_mathbb R:=Notimes_mathbb Zmathbb R$ is the dual of $M_mathbb R$, i.e., I need to show that $N_mathbb R:=hom_mathbb Z(M,mathbb Z)otimes_mathbb Zmathbb Rconghom_mathbb R(M_mathbb R,mathbb R)$.

Any hint will be helpful.

Thank you.

vector-spaces commutative-algebra abelian-groups

share|cite|improve this question

asked Jul 26 at 13:05

2015

1,2881521

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $M$ be a lattice i.e., $M$ is an abelian group and $Mcongmathbb Z^n$ for some $n$. Let $N:=hom_mathbb Z(M,mathbb Z)$. Let $M_mathbb R$ denotes the $mathbb R$-vector space $Motimes_mathbb Zmathbb R$. Then clearly $M_mathbb Rcongmathbb R^n$.

I need to show that $N_mathbb R:=Notimes_mathbb Zmathbb R$ is the dual of $M_mathbb R$, i.e., I need to show that $N_mathbb R:=hom_mathbb Z(M,mathbb Z)otimes_mathbb Zmathbb Rconghom_mathbb R(M_mathbb R,mathbb R)$.

Any hint will be helpful.

Thank you.

vector-spaces commutative-algebra abelian-groups

share|cite|improve this question

asked Jul 26 at 13:05

2015

1,2881521

Let $M$ be a lattice i.e., $M$ is an abelian group and $Mcongmathbb Z^n$ for some $n$. Let $N:=hom_mathbb Z(M,mathbb Z)$. Let $M_mathbb R$ denotes the $mathbb R$-vector space $Motimes_mathbb Zmathbb R$. Then clearly $M_mathbb Rcongmathbb R^n$.

I need to show that $N_mathbb R:=Notimes_mathbb Zmathbb R$ is the dual of $M_mathbb R$, i.e., I need to show that $N_mathbb R:=hom_mathbb Z(M,mathbb Z)otimes_mathbb Zmathbb Rconghom_mathbb R(M_mathbb R,mathbb R)$.

Any hint will be helpful.

Thank you.

vector-spaces commutative-algebra abelian-groups

share|cite|improve this question

asked Jul 26 at 13:05

2015

1,2881521

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 13:05

2015

1,2881521

asked Jul 26 at 13:05

2015

1,2881521

asked Jul 26 at 13:05

2015

1,2881521

1,2881521

add a comment | 

add a comment | 

2 Answers
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4
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To construct such an isomorphism, you could start by constructing an $BbbR$-linear map $M_BbbR longrightarrow BbbR$ for every pure tensor $fotimes cinoperatornameHom_BbbZ(M,BbbZ)otimes_BbbZBbbR$, where $finoperatornameHom_BbbZ(M,BbbZ)$ and $cinBbbR$. There are not many sensible options for such a construction. Then show that this extends to an $BbbR$-linear map
$$operatornameHom_BbbZ(M,BbbZ),otimes_BbbZBbbR longrightarrow operatornameHom_BbbR(M_BbbR,BbbR).$$
Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $BbbR$.

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answered Jul 26 at 13:38

Servaes

20k33484

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Note that $M cong mathbbZ^n$ is isomorphic to $N cong hom_mathbbZ(mathbbZ^n,mathbbZ)$ as $mathbbZ$-modules, and $M_mathbbR cong mathbbR^n$ is isomorphic to $hom_mathbbR(mathbbR^n,mathbbR)$ as $mathbbR$-modules. So, the result follows.

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edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

To construct such an isomorphism, you could start by constructing an $BbbR$-linear map $M_BbbR longrightarrow BbbR$ for every pure tensor $fotimes cinoperatornameHom_BbbZ(M,BbbZ)otimes_BbbZBbbR$, where $finoperatornameHom_BbbZ(M,BbbZ)$ and $cinBbbR$. There are not many sensible options for such a construction. Then show that this extends to an $BbbR$-linear map
$$operatornameHom_BbbZ(M,BbbZ),otimes_BbbZBbbR longrightarrow operatornameHom_BbbR(M_BbbR,BbbR).$$
Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $BbbR$.

share|cite|improve this answer

answered Jul 26 at 13:38

Servaes

20k33484

add a comment | 

up vote
4
down vote



accepted

To construct such an isomorphism, you could start by constructing an $BbbR$-linear map $M_BbbR longrightarrow BbbR$ for every pure tensor $fotimes cinoperatornameHom_BbbZ(M,BbbZ)otimes_BbbZBbbR$, where $finoperatornameHom_BbbZ(M,BbbZ)$ and $cinBbbR$. There are not many sensible options for such a construction. Then show that this extends to an $BbbR$-linear map
$$operatornameHom_BbbZ(M,BbbZ),otimes_BbbZBbbR longrightarrow operatornameHom_BbbR(M_BbbR,BbbR).$$
Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $BbbR$.

share|cite|improve this answer

answered Jul 26 at 13:38

Servaes

20k33484

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

To construct such an isomorphism, you could start by constructing an $BbbR$-linear map $M_BbbR longrightarrow BbbR$ for every pure tensor $fotimes cinoperatornameHom_BbbZ(M,BbbZ)otimes_BbbZBbbR$, where $finoperatornameHom_BbbZ(M,BbbZ)$ and $cinBbbR$. There are not many sensible options for such a construction. Then show that this extends to an $BbbR$-linear map
$$operatornameHom_BbbZ(M,BbbZ),otimes_BbbZBbbR longrightarrow operatornameHom_BbbR(M_BbbR,BbbR).$$
Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $BbbR$.

share|cite|improve this answer

answered Jul 26 at 13:38

Servaes

20k33484

To construct such an isomorphism, you could start by constructing an $BbbR$-linear map $M_BbbR longrightarrow BbbR$ for every pure tensor $fotimes cinoperatornameHom_BbbZ(M,BbbZ)otimes_BbbZBbbR$, where $finoperatornameHom_BbbZ(M,BbbZ)$ and $cinBbbR$. There are not many sensible options for such a construction. Then show that this extends to an $BbbR$-linear map
$$operatornameHom_BbbZ(M,BbbZ),otimes_BbbZBbbR longrightarrow operatornameHom_BbbR(M_BbbR,BbbR).$$
Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $BbbR$.

share|cite|improve this answer

answered Jul 26 at 13:38

Servaes

20k33484

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 13:38

Servaes

20k33484

answered Jul 26 at 13:38

Servaes

20k33484

answered Jul 26 at 13:38

Servaes

20k33484

20k33484

add a comment | 

add a comment | 

up vote
0
down vote

Note that $M cong mathbbZ^n$ is isomorphic to $N cong hom_mathbbZ(mathbbZ^n,mathbbZ)$ as $mathbbZ$-modules, and $M_mathbbR cong mathbbR^n$ is isomorphic to $hom_mathbbR(mathbbR^n,mathbbR)$ as $mathbbR$-modules. So, the result follows.

share|cite|improve this answer

edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

add a comment | 

up vote
0
down vote

Note that $M cong mathbbZ^n$ is isomorphic to $N cong hom_mathbbZ(mathbbZ^n,mathbbZ)$ as $mathbbZ$-modules, and $M_mathbbR cong mathbbR^n$ is isomorphic to $hom_mathbbR(mathbbR^n,mathbbR)$ as $mathbbR$-modules. So, the result follows.

share|cite|improve this answer

edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note that $M cong mathbbZ^n$ is isomorphic to $N cong hom_mathbbZ(mathbbZ^n,mathbbZ)$ as $mathbbZ$-modules, and $M_mathbbR cong mathbbR^n$ is isomorphic to $hom_mathbbR(mathbbR^n,mathbbR)$ as $mathbbR$-modules. So, the result follows.

share|cite|improve this answer

edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

Note that $M cong mathbbZ^n$ is isomorphic to $N cong hom_mathbbZ(mathbbZ^n,mathbbZ)$ as $mathbbZ$-modules, and $M_mathbbR cong mathbbR^n$ is isomorphic to $hom_mathbbR(mathbbR^n,mathbbR)$ as $mathbbR$-modules. So, the result follows.

share|cite|improve this answer

edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 23:45

edited Jul 26 at 23:45

edited Jul 26 at 23:45

answered Jul 26 at 14:27

dsw

963

answered Jul 26 at 14:27

dsw

963

answered Jul 26 at 14:27

dsw

963

963

add a comment | 

add a comment | 

 

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