Why torus is a 3-dimension object Torus is obtained from $\mathbb S^1\times \mathbb S^1$, each $\mathbb S^1 \subset \mathbb R^2$, then why torus isn't a 4-dimension object, but is a 3-dimension object?
Generally, what is the criteria that an object, obtained from product topology, has to be the same dimension as the product, or lower than the dimension of the product? Is it possible to obtain an object with dimension that is higher than the product?
Thank you!
 A: When you make the product of spaces their dimensions get summed. The product of two 1-dimensional objects in a 2-dimensional euclidean space is a 2-dimensional object in 4-dimensional euclidean space.
Then, in the case of the flat torus, it is possible to deform this 2-dimensional object (the flat torus in 4-dimensional euclidean space) to fit into a 3-dimensional euclidean space.
A: It is actually a two-dimensional object (in the sense it has dimension two as a manifold) which we often imagine living in a three-dimensional world but in your case, it actually sits naturally in $\mathbb{R}^4$ (the four-dimensional world). Note that topologically $S^1 \times S^1$ looks like the boundary of a "donut", not including the interior points. To describe a point in $S^1 \times S^1$ (or on the boundary of a donut), we need locally two parameters and in this sense it is a two-dimensional object.
In general, if $M,N$ are manifolds of dimension $m,n$ respectively, the product $M \times N$ will be a manifold of dimension $m + n$.
A: The torus itself is a $2$-dimensional creature. If you want to embed it in Euclidean space, it is most natural to do so in $4$-dimensional space, although it's possible to embed it in $3$-dimensional space as well. You lose some of the symmetry, though, as in the $\Bbb R^3$ version, going around the torus in each of the two cardinal directions (as indicaded in this picture) looks very different. They two directions look exactly the same in $\Bbb R^4$.
