How #P is easier than PSPACE?

Question

It is said that #P is not harder than PSPACE. This way, how can Alternating Turing machine solve #SAT in polynomial time?

And if it can, seems, that it takes at least $O(n^4)$ time which is more time than needed for TQBF.

Answer 1

This way, how can Alternating Turing machine solve #SAT in polynomial time?

An alternating Turing machine can only solve decision problems, so that questions makes sense for the decision variant of $\text{#SAT}$:

$$\text{Input: Boolean Formula $F$, $k \in \mathbb N$. Question: Are there at least $k$ solutions for $F$?}$$

This decision variant is in $\text{P}^\text{#P}$ ($\text{P}$ with $\text{#P}$ oracle), because you only need to ask the oracle how many solutions there are and compare the answer to $k$. It is proven that: $$\text{P}^\text{#P} \subseteq \text{PSPACE} = \text{AP}$$ So there exists indeed an alternating Turing machine solving the decision variant of $\text{#SAT}$ in polynomial time.

And if it can, seems, that it takes at least $\mathcal O(n^4)$ time which is more time than needed for TQBF.

Reference? That still wouldn't mean that the decision variant of $\text{#SAT}$ is harder than $\text{TQBF}$.

The term $A$ is not harder than $B$ is in this context defined as $A \le^p_m B$. That means there exists a polynomial time reduction from $A$ to $B$.

$\text{TQBF}$ is $\text{PSPACE}$-complete so from every decision problem in $\text{PSPACE}$ there exists a polynomial time reduction to $\text{TQBF}$.

Calculating the polynomial time reduction from the decision variant of $\text{#SAT}$ to $\text{TQBF}$ might already use more time than actually solving the resulting $\text{TQBF}$-instance (on an alternating Turing machine).