- #1
binbagsss
- 1,278
- 11
Hi,
Excuse me this is probably a really stupid question but I ask because I thought that the definition of the dimension of a space is the number of elements in the basis.
Now I have a theorem that tells me that
## dim M_{k} = [k/12] + 1 if k\neq 2 (mod 12)
=[k/12] if k=2 (mod 12) ##
for ## k >0 even ##
So e.g ## dim (M_4) = dim (M_10) = etc =dim(14) ## and ## dim(M_{12})=2 ##
Where ## M_{k} ## denotes a modular form of weight k
I also have proposition ( ring of mod forms) that : ## E_4 ## & ## E_6 ## the Eisenstein series form a basis for ## M_{k} ##, that is every modular form can be written uniquely as a polynomial in ## E_4 ## & ## E_6 ##My question
If ## E_4 ## & ## E_6 ## are a basis for the modular form then isn't this saying that the dimension of any modular form is 2?
So ## dim (M_k) ## is the number of unique matrices with given weight ## k ## in ## SL_2(z) ## right? So Obviously from the above this isn't correct, so I'm unsure on what the definition of 'basis' is here or...?
Many thanks in advance
Excuse me this is probably a really stupid question but I ask because I thought that the definition of the dimension of a space is the number of elements in the basis.
Now I have a theorem that tells me that
## dim M_{k} = [k/12] + 1 if k\neq 2 (mod 12)
=[k/12] if k=2 (mod 12) ##
for ## k >0 even ##
So e.g ## dim (M_4) = dim (M_10) = etc =dim(14) ## and ## dim(M_{12})=2 ##
Where ## M_{k} ## denotes a modular form of weight k
I also have proposition ( ring of mod forms) that : ## E_4 ## & ## E_6 ## the Eisenstein series form a basis for ## M_{k} ##, that is every modular form can be written uniquely as a polynomial in ## E_4 ## & ## E_6 ##My question
If ## E_4 ## & ## E_6 ## are a basis for the modular form then isn't this saying that the dimension of any modular form is 2?
So ## dim (M_k) ## is the number of unique matrices with given weight ## k ## in ## SL_2(z) ## right? So Obviously from the above this isn't correct, so I'm unsure on what the definition of 'basis' is here or...?
Many thanks in advance