Why use $-\dim(|A|)$ as target entropy in soft actor critic

Question

I am trying to implement Soft Actor-Critic with automatic Entropy tuning.

One thing I noticed is that the authors, and also the majority of implementations of SAC use $-\text{dim}(|\mathcal{A}|)$ as entropy target $\bar{\mathcal{H}}$ in the loss function of $\alpha$, and I am confused where this exact value came from. I have read the answer to this question, but I am still not fully convinced.

I do understand the intuition behind setting $\bar{\mathcal{H}}$ proportional to $\text{dim}(|\mathcal{A}|)$. Let's say, for example, we consider the target distribution of $\text{tanh}(\bf{u})$ to be $\text{Uniform}(-1, 1)$ for each dimension (I am not sure if this is a good assumption though), then the entropy target should be $\text{dim}(|\mathcal{A}|) \ln(2)$. But I am not sure why the majority of the implement uses exactly $-\text{dim}(|\mathcal{A}|)$.

Is there any specific reason for this setting? Or, are there any papers or articles about how the target entropy setting affects the performance of SAC?

Edit:

Just in case, the loss function of $\alpha$ I am referring to is Eq18:

$$ J(\alpha) = \mathbb{E}_{\bf{a}_t \sim \pi_t} [- \alpha \log \pi_t(\bf{a}_t | \bf{s}_t) - \alpha \bar{\mathcal{H}}] $$

Answer 1

If you are minimizing a loss function $f$, and there is a function $g$ that is proportional to $f$ something like this
$$ g = k \cdot f $$ then by minimizing $f$ you are also minimizing $g$.

It also extends to all transformations of f that do not change the ordering. If for a transformation $h(f)$ if you were to sort both $f$ and $h(f)$ and the order of the result is identical, then minimizing on $f$ is the same as minimizing on $h(f)$. Its said a little awkwardly, sorry.

Positive examples:

lets say that $f=\operatorname{abs}(x)$, you could minimize $h(f)=f^2$ and get the minimum would be the same.

lets say that $f=\operatorname{abs}(x)$ and $x$ is constrained to $0<x$. you could minimize $h(f)=\log(f)$ and get to the same location.

Negative examples (how to do it wrong):

lets say that $f=\operatorname{abs}(x)$ and $x$ is NOT constrained to $0<x$. you could minimize $h(f)=\log(f)$ and you would run into the discontinuity at $x=0$ and your optimization would fail.