Calculating the tangent space of the manifold of probability measures on a finite set Let $I$ be a finite set, $\mathcal{F}(I):= \{I\to\mathcal{R}\}$ the vector space of functions on $I$ with basis $e_i$, where $e_i(i)=1, e_i(j) = 0 (i \neq j)$. Let $\mathcal{S}(I)$ be the corresponding dual space and $\{\delta^i\}_i$ the respective dual basis.
Consider $\mathcal{P}_+:= \{ \sum_{i\in I} \mu_i \delta^i  \mid \mu_i>0, \sum\mu_i=1 \}$. In the book that I'm reading the authors say the tangent space in a point $\mu \in \mathcal{P}_+$ is obsviously $\mathcal{S}_0(I):=\{\sum_{i \in I} \mu_i \delta^i \mid \sum \mu_i = 0\}$. Why is that ? I think in order to calculate the tangent space we would calculate the image of the differential of $\varphi: (\mu_1,...,\mu_{s-1})\to \sum_{i=1}^{s-1}\mu^i\delta^i+(1-\sum_{i=1}^{s-1}\mu_i)\delta^s$, but I'm not sure how to do this. Can some one help? Of course $\mathcal{S}_0(I)$ has the correct dimension and so it is at least isomorphic to the tangent space but whats special about it ?

One of the several characterizations of the tangent space is the span of all tangent vectors of curves passing through a point. Let $\gamma$ be a smooth curve in $\mathcal{P}_+$ passing through $\mu$ represented by $\gamma(t)=\sum_{i\in I}\nu_i(t)\delta^i$ s.t. $\nu_i(0)=\mu_i$. Since we have $\sum_{i\in I}\nu_i(t)=1$ for all $t$, we may differentiate this to obtain
$$0=\frac{d}{dt}\sum_{i\in I}\nu_i(t)=\sum_{i\in I}\nu_i'(t),$$
which holds for $t=0$ in particular. The claim follows.
