# Thread: dimension of the vector space spanned over the rationals, and odd zeta values.

1. ## dimension of the vector space spanned over the rationals, and odd zeta values.

Hi guys,

I'm reading a fascinating paper on the infinitude of zeta values of the form $\zeta(2s+1)$. Whilst I understand the whole idea of the proof, and a fair bit of the detail, there's one vital part that's bugging me. The paper starts off with the following line: "We provide a lower bound for the dimension of the vector space spanned over the rationals by 1 and by the values of the Riemann Zeta function at the first n odd integers."

That is, for every $\epsilon > 0$, there exists an integer $N(\epsilon)$ such that if $n > N(\epsilon)$,

$
\dim_{\mathbb{Q}}(\mathbb{Q} + \mathbb{Q}\zeta(3) + \ldots + \mathbb{Q}\zeta(2n+1)) \geq \frac{1-\epsilon}{1+\log(2)}\log(n)$

I'm a little uneasy with this notation, and I find it slightly confusing. Is there any other way you can put this?

2. Here's a simpler statement: "The numbers 1, √2, e and π are linearly independent over the rationals." What that means is that the only rational numbers a, b, c, d such that a + b√2 + ce + dπ = 0 are a=b=c=d=0. To put it another way, the vector space over the rationals spanned by those four numbers has dimension 4. In symbols, $\dim_{\mathbb{Q}}(\mathbb{Q} + \mathbb{Q}\sqrt2 + \mathbb{Q}e + \mathbb{Q}\pi) = 4$.

For the Riemann zeta function, a naive person (like me, for example) might guess that the values of $\zeta(2k+1)$ should all be linearly independent over the rationals. In that case, the space spanned by 1 and the first n such values would have dimension n+1. What this paper is asserting is that for sufficiently large n this space has dimension at least log(n) (up to a constant multiple). I imagine that this would be a tough result to prove.

3. Originally Posted by Opalg

I imagine that this would be a tough result to prove.
yes, it's not that easy! especialy for those who only understand english. here's the paper for whoever is interested in Theorie des Nombres!

4. Originally Posted by NonCommAlg
yes, it's not that easy! especialy for those who only understand english. here's the paper for whoever is interested in Theorie des Nombres!
I remember the excitement at the International Congress of Mathematicians in Helsinki in 1978, when Apéry's result on the irrationality of ζ(3) has just become known. Things have come a long way since then!

5. Originally Posted by NonCommAlg
yes, it's not that easy! especialy for those who only understand english. here's the paper for whoever is interested in Theorie des Nombres!
Ah, it's fantastically difficult, but its such a fascinating area. I think it was Monsieur Rivoal who also conjectured that one of $\{\zeta(5), \zeta(7), \ldots, \zeta(21) \}$ was irrational. Apéry's proof is out of this world! How on earth did he come up with that!? I still actually don't fully understand that proof, nor did (do?) quite a few Mathematicians come to think of it!

Opalg, thank you, that's demystified that notation entirely!

6. Originally Posted by Opalg
I remember the excitement at the International Congress of Mathematicians in Helsinki in 1978, when Apéry's result on the irrationality of ζ(3) has just become known. Things have come a long way since then!
this is also a nice and recent result (2001) due to Zudilin that at least one of $\zeta(5) , \zeta(7), \zeta(9), \zeta(11)$ is irrational.