The multivariate stable distribution is defined as follows (see Nolan 2013 or wikipedia). Let $\mathbb{S}$ be the unit sphere in $\mathbb R^d\colon \mathbb{S} = \{u \in \mathbb R^d\colon|u| = 1\}$. A random vector $X$ has a multivariate stable distribution, that is $X \sim S(\alpha, \Lambda, \delta)$, if the joint characteristic function of $X$ is:
$$\operatorname{E} \exp(i u^T X) = \exp \left\{-\int \limits_{s \in \mathbb S}\left\{|u^Ts|^\alpha + i \nu (u^Ts, \alpha) \right\} \, \Lambda(ds) + i u^T\delta\right\}$$
For $y\in\mathbb R$,
$$\nu(y,\alpha) =\begin{cases} -\mathbf{sign}(y) \tan(\pi \alpha / 2)|y|^\alpha & \alpha \ne 1, \\ (2/\pi)y \ln |y| & \alpha=1. \end{cases}$$
$\delta \in \mathbb R^d$ is the shift parameter and $\Lambda$ is a spectral measure of the distribution. $0<\alpha<2$ is the stability parameter.
My question is this. Could someone please explain to me (in simple, intuitive terms) what $u$ is? It appears throughout the first equation, but I don't know what it is. What is its relationship with $s$?
Nolan, John P., Multivariate elliptically contoured stable distributions: theory and estimation, Comput. Stat. 28, No. 5, 2067-2089 (2013). ZBL1306.65118.
Let me make an analogy to a characteristic function of a real valued random variable.
Given a random variable $X$, its characteristic function is the map $\varphi_X \colon \mathbb{R} \rightarrow \mathbb{C}$ defined as:
$$ \varphi_X(u) \, \colon = \mathbf{E} \left[ e^{ i u X} \right].$$
Note that since the expectation is taken over the variable $X$, this becomes a function of $u$ only. Characteristic functions are useful, since if you have two random variables $X, \, Y$ and you can show that
$$\varphi_X(u) = \varphi_Y(u), \qquad \forall \, x,\,y \, \in \mathbb{R}$$
then it follows that $X$ and $Y$ have the same distribution.
Now returning back to your context: in your case the variable does not take values in $\mathbb{R}$, but rather in $\mathbb{R}^d$. The equivalent form of a characteristic function for a variable $X \in \mathbb{R}^d$ is a function $\varphi_X \colon \mathbb{R}^d \rightarrow \mathbb{C}$, where now:
$$\varphi_X( u) \, \colon = \mathbf{E} \left[ e^{i u^T X} \right],$$
note that we can take the expectation as $u^T X = \sum_{i=1}^d u_i X_i \in \mathbb{R}$ (since expectations only make sense for `1-dimensional' variables).
As in the case of $d = 1$, if you can show that given two variables $X, Y$ taking values in $\mathbb{R}^d$ such that
$$\varphi_X(u) = \varphi_Y(u), \qquad \forall \, x, \, y \in \mathbb{R}^d$$
then it follows that the variables are equal in distribution.
Further Introduction to Characteristic Functions
In this I go into some more detail to motivate characteristic functions. I return to the scenario that $X$ is a real valued variable, and $\varphi_X \colon \mathbb{R} \rightarrow \mathbb{C}$.
A common question in probability / statistics is: given two random variables / sets of data, how do I know if they have the same distribution.
A first approach might be to check: do they have the same expectation (mean, if doing statistics)? So suppose we have variables $X, Y$ we might say
$$ \text{Is } \mathbf{E}[X] = \mathbf{E}[Y] \text{ true?}$$
If its not then we'd conclude that they're definitely do not have the same distribution.
But, is the converse true? If $X, \, Y$ have the same mean, do they have the same distribution? The answer is no.
Example
Suppose $X \sim N(0,1)$ (a normal distribution with mean $0$ and standard deviation $1$), and $Y \sim \text{Unif}[-1,1]$ (a uniform variable on the interval $[-1,1]$). Both variables have the property
$$ \mathbf{E}[X] = \mathbf{E}[Y] = 0$$
However they definitely don't have the same distribution (since $\mathbf{P}[X > 1] > 0$, whilst $\mathbf{P}[Y > 1] = 0$).
So we might then ask: "If two random variables have the same mean, and the same variance... Do they have the same distribution?". Again the answer to this is no.
Example
Suppose $X \sim N(1,1)$ and $Y \sim \text{Poi}(1)$ (a Poisson variable with mean $1$). Then
$$\mathbf{E}[X] = \mathbf{E}[Y] = 1, \qquad \text{Var}(X) = \text{Var}(Y) = 1$$
Similarly we can construct examples for any given integer $k \geq 0$ that show that
$$\mathbf{E}[ X^k] = \mathbf{E}[Y^k]$$
and yet $X$ and $Y$ do not have the same distribution. We call $\mathbf{E}[X^k]$ the $k-$th moment of $X$.
However... It turns out (and this requires proving) that if you can show that all moments of the distributions agree: then they have the same distribution. i.e. if
$$\mathbf{E}[X^k] = \mathbf{E}[Y^k], \qquad \forall \, k \geq 0$$
then $X \sim Y$.
How does this relate to the characteristic function?
A similar statement from real analysis says that a power series uniquely determines a function. So the insight is: if we create a power series with the moments of $X$ as coefficients, and then we find a variable $Y$ with the same power series... Then: uniqueness of power series implies they have the same coefficients, i.e. the same moments, and therefore have the same distribution!
So we construct the power series:
$$ f(u) = \mathbf{E}[X^0] + \mathbf{E}[X^1] u + \mathbf{E}[X^2] \frac{u^2}{2!} + \mathbf{E}[X^3] \frac{u^3}{3!} + \cdots$$
We introduce the parameter $u$ purely as an artificial way to make a function from a sequence of moments $\{\mathbf{E}[X^k]\}_{k \geq 0}$. Using linearity of expectations, and the convenience of dividing the moments by factorials,we get:
$$ f(u) = \mathbf{E}\left[ 1 + uX + \frac{u^2X^2}{2}+ \frac{u^3X^3}{3!} + \cdots \right] = \mathbf{E}[ e^{uX} ]$$
The function $f$ is almost the characteristic function (its missing the $i$), and is known as the moment generating function. The above argument therefore says that if we have two random variables $X,Y$ and they have the same moment generating function, then they have the same distribution.
Introduction of the parameter $i$, so $f(iu)$, changes very little: the claim is still true that $\varphi = f(iu)$ uniquely determines the distribution. Really the addition of $i$ is historic, in as much as $\varphi$ is equivalent to the Fourier transform in real analysis.
So in short: the moments of a random variable uniquely determine the distribution. Therefore, a power series with the moments as coefficients uniquely determines the distribution. The characteristic function is exactly that: a power series, in some parameter $u$, with the moments of a random variable as coefficients.
The value of $u$ itself is not of interest: $u$ is a way to encode a sequence of information (the coefficients), as a function over a continuous parameter.