Vtyjkyuk

The complex torus is the quotient group of $mathbbC^n$ over a lattice. What is intuition behind that causes us to call it "torus"?

Let's consider just $Bbb C^1$ for now, and let's let the lattice be the one generated by $1$ and $i$, so that $a + bi = c+di$ exactly when both $a-c$ and $b-d$ are integers; all other lattices are quite similar.

Now it should be clear that the lattice divides the plane into squares, and each square is identified with the single principal square whose vertices are $0, 1, i, 1+i$. For example, under the quotient operation, $ 3¾+ (2+pi)i$ is the same point as $ ¾+ (pi-3)i$. So we can forget the rest of the plane: the quotient operation turns it into a square.

Additionally, the quotient operation identifies the north and south edges of the square, because $ x$ and $x+i$ are identified for each real $x$. We can imagine that we have rolled the square into a tube shape and glued together the two edges. The tube's boundaries are two circles, the left-side one consisting of the points $iy$ for each $yin Bbb R$ and
the right-side one consisting of the points $ 1+iy$ for each such $y$.

Additionally, the quotient operation identifies the west and east edges of the square,
because $iy$ and $1+iy$ are identified for each real number $y$. We can imagine that we should take the tube and bend it around and glue together the two circles. This makes a torus.

(If we imagine doing this in a three-dimensional space, the geometry of the tube is stretched as we bend it, and distances are not preserved. But this is only an artifact of our three-dimensional brains, and you should ignore it. If done properly, in a space of four or more dimensions, we can join together the two circles with no stretching, and the resulting torus is perfectly flat.)

So for the case of $Bbb C^1$ we do indeed get something that behaves just like the torus $Bbb T^2 = S^1times S^1$. The correspondence is quite exact. Say that $xin S^1$ and $yin S^1$. Then we have $langle x,yrangle in S^1times S^1$. To which point of the $Bbb C^1$ quotient space does this correspond? To $x+iy$, obviously! (Observe also that $S^1$ itself is the quotient space of $[0,1]$ under the identification that glues together the points $0$ and $1$.)

Similarly, when we do the same thing for $Bbb C^n$ we do get a space homeomorphic to $Bbb T^2n$ which is a direct product of $2n$ copies of the circle.