quantum mechanics - Why is the dimension of the set separable states $dimmathcal H_1+dimmathcal H_2$?

Please can you help me to understand how the dimension of the set of separable states is $\dim \cal H_1 + \dim \cal H_2$?

This is the relevant passage:

So far, we have assumed implicitly that the system is made of a single component. Suppose a system is made of two components; one lives in a Hilbert space $\cal H_1$ and the other in another Hilbert space $\cal H_2$. A system composed of two separate components is called bipartite. Then the system as a whole lives in a Hilbert space $\cal H = \cal H_1 \otimes \cal H_2$, whose general vector is written as $$\left|\, \psi \right\rangle = \sum_{i,j} c_{ij} \left|\,e_{1,i}\right \rangle \otimes \left|\,e_{2,j}\right\rangle, \tag{2.29}$$ where $\{|\,e_{a,i}\rangle\}$ ($a=1,2$) is an orthonormal basis in $\cal H_a$ and $\sum_{i,j} |c_{ij}|^2 = 1$.

A state $|\,\psi \rangle \in \cal H$ written as a tensor product of two vectors as $|\,\psi \rangle = |\,\psi_1 \rangle \otimes |\,\psi_2\rangle$, ($|\,\psi_a\rangle \in \cal H_a$) is called a separable state or a tensor product state. A separable state admits a classical interpretation such as “The first system is in the state $|\,\psi_1\rangle$, while the second system is in $|\,\psi_2\rangle$.” It is clear that the set of separable states has dimension $\dim \cal H_1 + \dim \cal H_2$.

Answer

Note that the space of separable states is not a vector space, and in particular not a subspace of the total Hilbert space: the sum of two separable states is unlikely to be separable. So dimension here means something more general than vector space dimension.

Having said that, I would disagree with the author on his dimension! I would say that the space of (nonzero) separable states has dimension $\dim \mathcal{H_1}+\dim\mathcal{H_2}-1$.

To specify a separable state, we can supply an element of each of $\mathcal{H_1}$ and $\mathcal{H_2}$, which means $\dim \mathcal{H_1}+\dim\mathcal{H_2}$ complex numbers. However, there is a redundancy here, because we can change each by an overall scaling ($|\psi_1\rangle\mapsto\lambda|\psi_1\rangle, |\psi_2\rangle\mapsto\lambda^{-1}|\psi_2\rangle$) without changing the product state, which reduces the dimension by 1.

A couple of simple examples:

1) If $\mathcal{H_1}$ is 1-dimensional (completely trivial!), then all states are separable, and $\mathcal{H_1}\otimes\mathcal{H_2}\simeq\mathcal{H_2}$.

2) If both $\mathcal{H_1}$ and $\mathcal{H_2}$ are two-dimensional, we can write a state of $\mathcal{H_1}\otimes\mathcal{H_2}$ as a 2x2 matrix. The separable states have proportional columns/rows, so are exactly the same as matrices of determinant zero. If we exclude 0, this is a 3-dimensional submanifold.