Tensor Product of a Bra and a Ket What does one get if the take the tensor product of a bra and a ket, for instance, $\langle\uparrow \rvert \otimes \lvert \downarrow\rangle$?
What I mean it, what is this object? What does it act on? Or does it get acted on? Is it meaningful, like the tensor product of two kets is, etc?
 A: The confusion is coming from the fact that you're thinking in terms of the bra-ket physics notation without understanding how the underlying vector spaces are constructed.
"Kets" are vectors in a vector space, i.e. a set of objects on which vector-vector addition and vector-scalar multiplication is defined (for some field of scalars).  "Bra"s are covectors, aka one-forms, defined as linear functions from a vector space to its field of scalars.  They also form a vector space, and they exist even if we don't define an inner product on the set of kets.
Since the space of bras is a vector space, it can be tensored with another vector space such as the space of kets.  This is defined just like any other tensor product of two vector spaces (which is the Cartesian product, equipped with an intuitive definition of addition and multiplication).  You could write this down in two equivalent ways: $\vert{\uparrow}\rangle \otimes \langle{\downarrow}\vert$ or $\langle{\downarrow}\vert \otimes \vert{\uparrow}\rangle$.  They are structurally the same thing, just with a different convention for the ordering.  (Likewise, coordinates in 3D can be ordered $(x,y,z)$ or $(z,y,x)$ without changing anything.)
The first case is often abbreviated $\vert{\uparrow}\rangle \langle{\downarrow}\vert$ since there's no risk of confusing it with anything else. (Of course, that notation is a bit dangerous since it suggest we have a way of associating a bra $\langle x \vert$ with every ket $\vert x \rangle $, but that would be a mistake because we haven't defined an inner product.)
On the other hand, if you tried to write the second case as $\langle{\downarrow}\vert \vert{\uparrow}\rangle$ or $\langle{\downarrow}\vert{\uparrow}\rangle$, it could be misinterpreted as letting the bra (which is a linear functional from the space of kets to the scalars) act on the ket, which produces a scalar.  That's a very different mathematical object.  If the bras and kets have vector space dimension $N$, then the object  $\langle{\downarrow}\vert{\uparrow}\rangle$ has dimension 1 while the object  $\langle{\downarrow}\vert \otimes \vert{\uparrow}\rangle$ has dimension $N^2$.
