# Why is complex torus “torus”?

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

• Maybe not relevant but torus is used when "every coordinate loops over itself", like the meridian and parallel of a torus. – user65203 Aug 29 '18 at 10:06
• In the case of $n=1$, it is homeomorphic to a torus – user99914 Aug 29 '18 at 10:07
• The same reason why we call $\mathbb S^n$ the sphere, I guess. – user99914 Aug 29 '18 at 10:07
• According to wikipedia, it's because they are exactly, up to isomorphism, the complex manifolds whose underlying smooth structure is a torus. – Saucy O'Path Aug 29 '18 at 10:07
• Further to the point made by Yves, groups like SL($2$,$\mathbb{C}$) and SL($n+1$,$\mathbb{C}$) have a similar important subgroups isomorphic to, say, $\mathbb{C}^{*}$ so the name torus is applied there, too. – SoboKevSpace Aug 29 '18 at 10:10

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 $y\in \Bbb R$ 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 $\Bbb T^2 = S^1\times S^1$. The correspondence is quite exact. Say that $x\in S^1$ and $y\in S^1$. Then we have $\langle x,y\rangle \in S^1\times S^1$. 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.