What does "$\Sigma^{\frac12}$" mean in the context of variational auto encoders? I am studying about Variational Auto Encoders. I found the following equation. 

$$\mu +\Sigma^\frac12\odot\epsilon$$

I can imagine that $\Sigma$ is related to variance. What does the  $\Sigma^{1/2}$ mean? And how is the notation called? I can not find any good articles related to this.
 A: This literature is honestly not very well-notated in general. Let me try to clarify.
Usually a 1D normally distributed random variable $\widetilde{z}$ (dependent on some other variable $x$) is written like: $$ \widetilde{z}\sim\mathcal{N}(\mu(x),\sigma(x)^2) $$
where $\mu(x)\in\mathbb{R}$ is the mean and $\sigma(x)^2\in\mathbb{R}$ is the variance. An equivalent value can be sampled via $ \widetilde{z} = \mu(x) + \varepsilon\sqrt{\sigma(x)} $ where $\varepsilon\sim\mathcal{N}(0,1)$. Notice the square root is used to get the standard deviation.
Separately, a multivariate, normally distributed random variable is often written as:
$$ z\sim\mathcal{N}(\mu(x),\Sigma(x)) $$
where $z,\mu(x)\in\mathbb{R}^n$ and $\Sigma(x)\in\mathbb{R}^{n\times n}$. Note that $\Sigma(x)$ must be positive definite and symmetric in order to be a proper covariance matrix. In this case, we can sample a value via $z = \mu(x) + \sqrt{\Sigma(x)}\,\xi$, where $\xi\sim\mathcal{N}(0,I)$, $\xi\in\mathbb{R}^n$. Here, $\sqrt{\Sigma(x)}=\Sigma^{1/2}(x)\in\mathbb{R}^{n\times n}$ is the matrix square root, i.e. $\Sigma^{1/2}(x)\Sigma^{1/2}(x)=\Sigma(x)$. Note that one cannot in general compute this by taking the square root of the elements of the matrix (though one can for diagonal matrices)! This can be non-trivial (e.g. see here or here).
However, for VAEs and the reparameterization trick, in most cases the covariance matrix is defined to be diagonal. This tells us two things: (1) the matrix-vector product $\sqrt{\Sigma}\xi$ becomes a Hadamard product, i.e. in this case $ \sqrt{\Sigma}\odot\xi= \sqrt{\Sigma}\xi$, and (2) the matrix root becomes an element-wise root, i.e.
$\Sigma = \text{diag}(\Sigma_1,\ldots,\Sigma_n)\;\implies\;\Sigma^{1/2}=\text{diag}(\Sigma^{1/2}_1,\ldots,\Sigma^{1/2}_n)$. Thus, sampling a value in a VAE with diagonal covariance becomes:
$$ z = \mu(x) + \sqrt{\Sigma(x)}\,\xi = \mu(x) + \Sigma^{1/2}(x)\odot\xi $$
where you don't have to worry about matrix roots, because you can take the elementwise root! :)
