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I need to be sure that I'm not writing mistakes in my notes on topological spaces. So the question is very simple :

Is it true that the locally euclidian 3D space obtained by identify the opposite faces of a cube (periodic conditions) isn't simply connected ?

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    Do you understand why this 3-torus is a three-fold product of circles $S^1 \times S^1 \times S^1$? From there, we see that the fundamental group $\pi_1(S^1 \times S^1 \times S^1) \cong \pi_1(S^1)^3 \cong \mathbb{Z}^3$ is nontrivial. – 2017-02-11
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    I was trying to see it "visually". The straight line going through all that space is closing on itself (from the periodic conditions), and was wondering if I could deform it to a point. Apparently not, so this seems to show that the flat 3D torus isn't simply connected, right ? – 2017-02-11
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    Yes, it isn't simply connected. One of the generators of the fundamental group $\mathbb{Z}^3$, say $(1,0,0)$, corresponds to the (homotopy class of the) loop that goes in a straight line from the center of the cube, to one of the faces, and back to the center from the opposite face. Since this this is nonzero in the fundamental group, it represents a noncontractible loop. The other two generators $(0,1,0)$ and $(0,0,1)$ would correspond to the loops through the two other pairs of opposite faces of the cube. – 2017-02-11

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Yes, that is true. The closed curve that starts at the center of your cube, goes straight up through the horizontal side and keeps going up to reach the center again, cannot be contracted to a point.

The most concrete way to see this may be by considering the torus to be a quotient of $\mathbb R^3$. Every curve in the torus, together with one point in $\mathbb R^3$ that maps to the curve's starting point, determines exactly one curve in $\mathbb R^3$, and does so continuously. The curve I describe above corresponds to a curve in $\mathbb R^3$ that goes between different representative points of the center, and it it were homotopic to a point, the end point would need to move continuously from one representative of the center to the other, while still being a representative all the way, which is absurd.

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    So this space is still *locally* **euclidian**, right ? Is a *globally* euclidian space defined as *simply connected* ? – 2017-02-11
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    @Cham: I'm not sure "globally euclidean" is a term in actual use, but if it is to mean anything I think it would be "homeomorphic to $\mathbb R^n$ for some $n$" -- and all _those_ spaces happen to be simply connected. – 2017-02-11
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    Thanks. I think it's all clear. Hopefully, I did not wrote mistakes in my notes. – 2017-02-11
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    Actually, there's a simple sentence in my notes that I need to be sure is true : refering to the flat 3D torus, I wrote "*This euclidian 3D space isn't simply connected*". The term "euclidian" is what makes me a bit nervous. I hope it is clear that it's actually meaning "locally euclidian 3D space". – 2017-02-11