What to $t\wedge s$ mean when $t$ and $s$ are just scalars?

Question

In my experience $\wedge$ has something to do with the outer product, but I am not sure what it means when $t$ and $s$ are not vectors and the book I am reading does not explain it. I thought maybe it means $min\{s,t\}$ but before the book used it it had already specified that $s<t$, so we already know a priori what the minimum is.

Answer 1

I was able to find a free (legal) preview of the book by searching for it on Amazon -- the preview includes page 25. I see why you are confused. I don't know if this is the right answer, but here is how I am understanding it.

They assumed $t > s$ when they defined the kernel $K_{H}(t,s)$. And then they defined another function afterward:

$R_{H}(t,s) = \displaystyle \int_{0}^{t \land s} K_{H}(t,u)K_{H}(s,u) \,du$.

I think that the formula they gave for $K_{H}(t,s)$ only holds if, when $t$ is fixed, the input $s$ is smaller than $t$. And you can see in the definition of $R_{H}(t,s)$ that it makes sense. In that definition, for fixed $t, s$, since we are integrating with respect to $u$ over $[0, t \land s]$, then each $u$ is already smaller than $t$, so the first factor in the product, $K_{H}(t,u)$ makes sense by the formula they gave. Also, each $u$ is smaller than $s$, so the second factor in the product, $K_{H}(s,u)$, make sense by the formula they gave.

So, if $t$ is fixed, then for each $s < t$, $K_{H}(t,s) = $ that formula they gave. Then for any $s, t$ (no matter which is bigger), $R_{H}(t,s)$ is the function that sends $(t,s)$ to the integral (w.r.t. $u$) over $[0, t\land s]$ of the product $K_{H}(t,u)K_{H}(s,u)$, and each of the factors in this product are defined by the $K_{H}$ formula since $u$ is always smaller than $t$ and $s$ (since it's smaller than $t \land s$ if $\land$ is the min symbol).