Is $p \vee q \leq p+q$ for $p,q$ projections?

Question

I am wondering if $p \vee q \leq p+q$ for $p,q$ projections acting on some Hilbert space $H$.

In particular, I wonder if the set of finite trace projections is upwards directed with the usual ordering of positive elements.

Answer 1

To your first question, let $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$ Then $$ p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix}, $$ so $$ p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}, $$ which is not positive nor negative, so $p\vee q$ and $p+q$ are not comparable.

To your second question: yes, the set of projections is a lattice, i.e. in general $$ p\wedge q\leq p,q\leq p\vee q. $$ And, if $p$ and $q$ are finite-rank (equivalently, finite-trace), then so is $p\vee q$.