Modular Forms, Dimension, Valence Formula

[binbagsss]

Homework Statement

What is the dimension of $M_{24}$?

Homework Equations

attached

The Attempt at a Solution

[/B]
I am confused what the (mod 12) is referring to- is it referring to the $[k/12]$ where the square brackets denote an equivalent class and the $k \equiv 2$ / $k \notequiv 2$ or just the $[k/12]$?

I am confused because I thought $k \equiv 2$ (mod 12) only when $k=24$, so for the dimension $M_2$ we would need to look at the top definition, however clearly the bottom has been used, which makes me think that the '(mod $12$)' is only referring to the square brackets?

In which case for $M_{24}$ I need to look at the top line and conclude $dim M_{24}=3$, however if (mod 12) is referring to both then I need to look at the bottom line and conclude $dim M_{24}=2$, however in this case it makes no sense how we have got $dim M_2=0$

Thanks .

 

[fresh_42]

Homework Statement

What is the dimension of $M_{24}$?

Homework Equations

attachedView attachment 195345

The Attempt at a Solution

[/B]
I am confused what the (mod 12) is referring to- is it referring to the $[k/12]$ where the square brackets denote an equivalent class and the $k \equiv 2$ / $k \notequiv 2$ or just the $[k/12]$?

I am confused because I thought $k \equiv 2$ (mod 12) only when $k=24$, so for the dimension $M_2$ we would need to look at the top definition, however clearly the bottom has been used, which makes me think that the '(mod $12$)' is only referring to the square brackets?

In which case for $M_{24}$ I need to look at the top line and conclude $dim M_{24}=3$, however if (mod 12) is referring to both then I need to look at the bottom line and conclude $dim M_{24}=2$, however in this case it makes no sense how we have got $dim M_2=0$

Thanks .

$24 \equiv 0 \operatorname{mod}12$
$2 \equiv 2 \operatorname{mod}12$

 

[binbagsss]

$2 \equiv 2 \operatorname{mod}12$

So
$4 \equiv 4 \operatorname{mod}12$

So from the definition above $dim M_{4} =[k/12]=[4/12]$;

how is $[4/12]$ 1? Isn't this zero too? what do the square brackets denote.

E.g $14\equiv 2$ (mod 12) so am I using the original $k$ : $[14/12]$ or $[2/12]$?

 

[binbagsss]

S

E.g $14\equiv 2$ (mod 12) so am I using the original $k$ : $[14/12]$ or $[2/12]$?

Oh it doesn't matter, [ ] denote equivalent classes, so it's 'the remainder of the division' which is $2$ in both of these cases?

Can I just test my understanding here- is $dim M_{28}=[k/12]+1=5$?

 

[binbagsss]

Oh it doesn't matter, [ ] denote equivalent classes, so it's 'the remainder of the division' which is $2$ in both of these cases?

Can I just test my understanding here- is $dim M_{28}=[k/12]+1=5$?

No I'm lost $dim M_{12}=2$ but I am getting:

$12 \equiv 0$ mod 12, so I'm looking at $[k/12]+1$, if these [ ] denote equivalent classes a number divisible by 12 is represented by the element $0$ so I get $0+1=1$...

Unless these [ ] square brackets denote taking the integer or something? what do these square brackets mean? thanks.

 

[Dick]

No I'm lost $dim M_{12}=2$ but I am getting:

$12 \equiv 0$ mod 12, so I'm looking at $[k/12]+1$, if these [ ] denote equivalent classes a number divisible by 12 is represented by the element $0$ so I get $0+1=1$...

Unless these [ ] square brackets denote taking the integer or something? what do these square brackets mean? thanks.

The brackets are the floor function. The greatest integer less than or equal to the quotient.