# A Modular forms, dimension and basis confusion, weight mod 1

1. Nov 23, 2016

### binbagsss

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...?

2. Nov 23, 2016

### zinq

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dimMk = [k/12] + 1ifk ≠ 2(mod12) = [k/12]ifk = 2(mod12)

for k > 0even
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binbagsss, your theorem is very hard to read, partly because all the characters are jammed together without spaces.

Also: Mk does not denote "a modular form of weight k". It denotes the vector space over the complex numbers of modular forms of weight k.

Then the correct statement is as follows:

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Let Mk denote the complex vector space of modular forms of weight k.

Let k ≥ 0 be odd. Then Mk = 0.

Let k ≥ 0 be even. Then

[k/12] + 1, k ≠ 2 (mod 12)​
dim(Mk) = {
[k/12], k = 2 (mod 12)​
-----

You wrote:
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My question

If E4 & E6 are a basis for the modular form then isn't this saying that the dimension of any modular form is 2?
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But a modular form is not a vector space. It is the space Mk of modular forms of a given weight k that is a vector space (over ). So a modular form does not have a dimension; only a vector space of modular forms does.

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So dim(Mk) is the number of unique matrices with given weight k in SL2(z) right?
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A modular form is not a matrix! And what do you mean by the word "unique"? I suggest you go over the definition of a modular form, and what "weight k" means. And what the dimension of a vector space means.

Also: I am guessing that by "E4" and "E6" you mean the modular forms that are typically denoted by G4 and G6, or in more generality:

G2k(τ) = ∑ 1/(mτ + n)2k,​

where τ is assumed to lie in the upper half-plane of ℂ, and the summation is over all pairs (m, n) of integers with (m, n) ≠ (0,0).

Then it can be shown that for k ≥ 2, G2k is a modular form of weight 2k.

Therefore, G4 and G6 are modular forms of different weights, so would not even lie in the same vector space Mk for any k.

3. Nov 23, 2016

### binbagsss

Right okay thanks, a bunch of mistakes I see, makes more sense now I think though, so...

Every modular form for $SL2(Z)$ can be expressed uniquely as a polynomial in $E4(τ )$ and $E6(τ )$. That is, ${E ^a _4 E_ b ^6 ; 4a + 6b = k}$ forms a basis for $Mk(Γ)$

This is in my lecture notes, given as a proposition, copied from them

So a modular form is a function with certain properties etc, so to talk of a single function having dimension makes no sense, I see that now, thanks,

The space of modular forms of different weights is the vector space, so that has dimension, fine

But I thought basis and dimension go together? One definition of the dimension of a space is the number of elements in the basis right? So isn't the term 'basis' used sort of as slang in the proposition above? Thanks.

4. Nov 23, 2016

### zinq

After fixing the mistakes in what you wrote, I get:

"Every modular form for SL2(Z) can be expressed uniquely as a polynomial in E4(τ) and E6(τ).

"That is, E4a E6b for 4a+6b=k forms a basis for Mk(Γ)"

It's hard to know what Γ means without a definition, but I will assume it means SL2(Z).

What this means is: For each k, the set

{E4a E6b | a and b are positive integers with 4a+6b=k}​

of modular forms forms a basis of Mk(Γ) over the complex numbers.

So: There is no slang occurring here. (Advanced math is difficult enough that people try to avoid using terms to mean other than their standard meaning!)

5. Nov 24, 2016

### binbagsss

Sorry no I'm still confused, for a specified $k$, so then isn't the number of elements in the basis equal to the dimension of the space? => 2, so I'm back at my initial question, with terminology corrected...thanks

Since dimension of $M_{k}$, for a given $k$ is something like the number of non co-linear functions that are modular forms of weight k, is this ok?

So a basis for $M_{k}$ is the minimum number of these functions that are linearly independent such that every modular form in that space, of a given weight k, can be expressed as a linear combination in terms of them?

So the proposition is saying that $E_{4}$ and $E_{6}$ do it, so is it not saying the basis consists of two elements?

Last edited: Nov 24, 2016
6. Nov 24, 2016

### zinq

Yes, the dimension of a vector space is the number of elements in any basis of it.

No, this is not saying that E4 and E6 are the two elements of a basis. First of all — as we said — they are not in the same vector space Mk.

This is the key statement:

{E4a E6b | a and b are positive integers with 4a+6b=k}.

This says that if you take the set of positive integers a and b for which 4a + 6b = k and create the modular form E4a E6b for each such a and b, then those modular forms are a basis for Mk. (That is, E4 raised to the power a, times E6 raised to the power b.)

binbagsss, I am pretty sure it would benefit you to review the fundamental definition of what it means to say that a subset of a vector space is a basis for it. Note that in this case the vectors of our vector space are the modular forms. Although modular forms from the vector spaces Mk for various k can be multiplied by each other, that has nothing to do with the structure of a vector space, which does not include the concept of multiplying vectors!

Last edited: Nov 24, 2016
7. Nov 25, 2016

### binbagsss

So a subset of a vector space is a basis for that vector space if any vector can be represented as a linear combination of them uniquely, and the dimension of a vector space V, denoted dim V, is the cardinality of its bases, which if finite is equal to the number of elements

this is just what i thought before, and I'm still lost...?

I have this picture is this correct:

- Choose $E_{4}^{a}$ , $E_{2}^{b}$ , these two vectors are the subset you are referring to, where the relevant vector space it is a subset of is $M_{k}$ s.t $a+b=k$ right ?

What is it saying then?

8. Nov 27, 2016

### zinq

I wonder if the problem might be understanding the expression

E4a E6b.​

This means:

(E4 raised to the power a) times (E6 raised to the power b).​

You can then check that

E4a E6b

is a modular form of weight k, where

k = 4a + 6b.​

9. Nov 28, 2016

### binbagsss

no that's fine . its not E4 and E6 that are the basis but E4 raised to a , E6 raised to b ...

10. Nov 30, 2016

### zinq

"Its not E4 and E6 that are the basis but E4 raised to a , E6 raised to b ..."

It's {E4a times E6b} for just those a and b >= 1 such that 4a + 6b = k ... that form a basis for Mk over the complex numbers.

11. Dec 1, 2016

### binbagsss

[QUOTE="zinq, post: 5631872, member: 462505

It's {E4a times E6b} for just those a and b >= 1 such that 4a + 6b = k ... that form a basis for Mk over the complex numbers.[/QUOTE]

that's one element?

12. Dec 1, 2016

### binbagsss

that's one element?[/QUOTE]

one dimension for one function? just one function of f in Mk ?
Then the total dimension of the basis for a given k is the number of functions in Mk?

13. Dec 1, 2016

### zinq

It would be most helpful for you to review the definition of a vector space and of a basis. The vector space Mk of modular forms of weight k has as elements (called "vectors" in this case) the modular forms of weight k. Your questions seem to indicate an unfamiliarity with these basic concepts. The best thing you can do now is to review these concepts and see how they apply to my answers.