The invariant property of Kalman filter

I came a cross a property for Kalman filter known as invariant property. I could only find some information about it on a wikipedia article but I still struggle to understand it.

The property is that if $E[x_k-\hat{x}_k]=0$, where $x_k$ is the state vector at time $k$ and $\hat{x}_k$ is the estimate of $x_k$, then $$COV(x_k-\hat{x}_k)=COV(\hat{x}_k) \tag1\label{1}$$, where $COV(x_k-\hat{x}_k)$ corresponds to the updated estimate covariance.

I am puzzled by this because on one hand, \eqref{1} implies that $$trace(COV(x_k-\hat{x}_k))=trace(COV(\hat{x}_k))$$ and on other other hand, $trace(COV(x_k-\hat{x}_k))=MSE(\hat{x}_k)$, and hence$$MSE(\hat{x}_k)=trace(COV(\hat{x}_k))+bias\tag2\label2$$ and the $bias$ is some function of $E[x_k-\hat{x}_k]$ and it is zero if and only if the invariant holds. So my questions are:

1. Is \eqref{2} correct? If it is not then what is the implication of invariant property for $MSE(\hat{x}_k)$?
2. If \eqref{2} is correct, then is it minimized by setting the bias to zero?
3. If Kalman for linear Gaussian system is optimal in terms of minimizing MSE, doesn't this make Kalman filtering and unbiased estimator? (how could a Bayesian estimator be unbiased?)
4. Finally, if all the above are based on my wrong understanding of invariant property, could someone elaborate on it. In particular, I am interested to know its implications on MSE, for example could I say a Kalman with invariant is optimum in MSE sense and outperforms Kalman without invriant propert?