Piketty's explanation of elasticity of substitution (from his book Capital in the 21st century) I have some trouble following the explanation of the elasticity of substitution between capital and labor and its implications on p189.
Take this part:

The relevant question is whether the elasticity of substitution
between labor and capital is greater or less than one. If the
elasticity lies between zero and one, then an increase in the
capital/income ratio β leads to a decrease in the marginal
productivity of capital large enough that the capital share α = r × β
decreases (assuming that the return on capital is determined by its
marginal productivity).

Specifically, when he talks about that elasticity, does he mean something like (dL/L)/(dK/K) with L for labor and K for capital? Assuming that, how exactly does a higher elasticity translate in a lower fall in marginal product of capital?
Thanks
The following is from Thomas Piketty and Gabriel Zucman (2015, From Handbook of Income Distribution, Volume 2, Chapter 15, Part 15.5.3 which is hard to link to directly but get it here):

Take a CES production function
$$Y=F(K,L)=(a⋅K^{\frac{\sigma-1}{\sigma}}+(1−a)⋅L^{\frac{\sigma-1}{\sigma}})^{\frac{\sigma}{1-\sigma}}$$

Me: $\sigma$ is the elasticity of substitution. They have a small typo here with the outer exponent of $\frac{\sigma-1}{\sigma}$ instead of $\frac{\sigma}{1-\sigma}$, but I'm pretty sure that is wrong so I corrected it here. It doesn't change the rest of what I'm doing.
Back to Piketty and Zucman:

The rate of return is given by $$r = F_{K} = a \cdot  \beta^{\frac{-1}{\sigma}}$$ with $\beta = K/Y$.
The capital share is given by $$ \alpha = r\cdot\beta = a\cdot\beta^{\frac{\sigma-1}{\sigma}}$$

Me: Now take the partial derivative of $\alpha$ with respect to to $\beta$
$$ \frac{\partial \alpha}{\partial \beta} = a \cdot (1-1/\sigma) \cdot \beta ^{-1/\sigma}$$
By assumption $a$ is always positive. Because $K$ and $Y$ are always positive so is $\beta ^{-1/\sigma}$. That means the sign of $\alpha$ is the sign of $1-1/\sigma$. For $0&lt;\sigma&lt;1$ you can see that $1-1/\sigma &lt; 0$ and therefore $$\frac{\partial \alpha}{\partial \beta}  &lt; 0$$.