Suppose I have a standard flat torus that is the quotient of $\mathbb{R}^2$ by the lattice $\Lambda = (\vec{e}_1, \vec{e}_2)$ and a matrix $A$ that represents rotation by an irrational angle between 0$^o$ and 90$^o$. What would be the induced map on $H_1(T^2)$ of the diffeomorphism $\phi$ of $T^2$ that takes the image of the geodesic at $t$ determined by the vector $\vec{v}_1$ at $(\theta_1, \theta_2) = (0,0)$ to the image of the geodesic at $t$ determined by the vector $A\vec{v}_1$ at $(\theta_1, \theta_2) = (0,0)$? One can't just look at the number of times the geodesics $\gamma_1$ induced by $A\vec{e}_1$ and $\gamma_2$ induced by $A\vec{e}_2$, say, wrap around each circle, because they're not closed geodesics.
Map Induced on $H_1(T^2)$ by Rotation by Irrational Angle
algebraic-topologydifferential-topologyriemannian-geometry
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It's a little silly to post an answer right after posting the question but I thought about it more and it turns out that such an action is impossible.
The idea for arguing based on the Mostow paper is from
https://math.stackexchange.com/a/4315382/758507
Theorem C of
https://www.researchgate.net/publication/227067355_A_Structure_Theorem_for_Homogeneous_Spaces
states that a compact homogeneous space is a fiber bundle (with connected base and fiber) with base a compact Riemannian homogeneous space and fiber a compact solvmanifold unless it is the quotient of a Lie group by a cocompact lattice (what Mostow calls a HLG= homogeneous local group).
This reference classifies all compact 3d HLG
So $M$ is not a HLG. So it must be a bundle of a solvmanifold over a compact Riemannian homogeneous space.
So we move on to the bundle case:
The base of this bundle cannot have dimension 3 as $ M $ is not a compact Riemannian homogeneous space (since it has no transitive action by a compact group). For example because $ \pi_1(M) $ does not have finite commutator subgroup
Transitive action by compact Lie group implies almost abelian fundamental group
So we must have a nontrivial fiber bundle:
If the fiber is dimension 3 we still have a nontrivial fiber bundle since every compact solvmanifold (except the circle) is a fiber bundle with base a torus and fiber a compact nilmanifold, both of strictly lower dimension. (fun fact: a compact nilmanifold is exactly an iterated principal circle bundle)
Since we are in such a low dimension a bundle must be either $$ S^1 \to M \to \Sigma $$ where $ \Sigma $ is Riemannian homogeneous ($ T^2,S^2,\mathbb{R}P^2 $) or the bundle must be $$ \Sigma \to M \to S^1 $$ where $ \Sigma $ is a solvmanifold (so either torus $ T^2 $ or Klein bottle $ K $). Since $ S^1 $ and $ M $ are aspherical $ \Sigma $ cannot be $ S^2 $ or $ \mathbb{R}P^2 $. So the fiber bundle must be either $$ S^1 \to M \to T^2 $$ or $$ K,T^2 \to M \to S^1 $$ But both of these are impossible because LES homotopy gives that $ \pi_1(M) $ is the semidirect product of $ \pi_1 $ of the base by normal action of $ \pi_1 $ of the fiber. Abelianizing we have that $ H_1(M) $ is direct product of $ H_1 $ of the base with a quotient of $ \pi_1 $ of the fiber. See for example
https://mathoverflow.net/questions/35713/abelianization-of-a-semidirect-product
but $ H_1 $ of the base always has free rank at least 1 for all these bundles so that would imply $ H_1(M) $ has free rank at least 1. That rules out the Hantzsche-Wendt Manifold which has first Betti number $0$.
The idea for ruling out bundles using the fact that $ M $ has vanishing first Betti number is due to
This is not an actual answer, but this is too long for a comment.
Is this a flat torus? If so, how would I find that lattice Λ from the above-cited post?
If you have already shown that all sectional curvatures of your torus $(S^1\times S^1,g)$ vanish (which I haven't checked), then the metric $g$ is flat by definition. From the classification of flat tori, there exist a lattice $\Lambda \subset \Bbb R^2$ and an isometry $f \colon \Bbb R^2/\Lambda \to (S^1\times S^1,g)$. The metric on the left is the one obtained from the Euclidean metric on $\Bbb R^2$.
Here is how to construct the corresponding lattice. Let $\gamma_1$ (resp. $\gamma_2$) be the shortest (resp. second shortest) closed unit speed geodesic in $(S^1\times S^1,g)$. Let $p_0$ be the point where they intersect. Up to translating the time parameter, $\gamma_1$ and $\gamma_2$, we can assume that $\gamma_1(0) = \gamma_2(0) = p_0$. Now, up to replacing $\gamma_2(t)$ with $\gamma_2(-t)$, we can assume that $\langle \gamma'_1(0),\gamma'_2(0) \rangle \geqslant 0$. Let $u = \gamma'(0)$ and $v = \gamma'(0)$. Let $\Lambda = \Bbb Z u \oplus \Bbb Z v \subset T_{p_0}(S^1\times S^1) = \Bbb R^2$ (this identification is canonical). The exponential map $\exp_{p_0} \colon \Bbb R^2 \to (S^1\times S^1,g)$ is a local isometry, and descends as an isometry $f \colon \Bbb R^2 / \Lambda \to (S^1\times S^1,g)$.
Is there a diffeomorphism from this torus to the standard torus $\Bbb R^2/ \Bbb Z^2$ that carries the Riemannian metric from this torus onto the standard one?
This will be possible if and only if $|u|=|v| = 1$, and $\langle u,v\rangle = 0$, i.e. if $\{u,v\}$ is an orthonormal basis of $T_{p_0}(S^1\times S^1)$. This is because two flat tori $\Bbb R^2 / \Lambda_1$ and $\Bbb R^2 / \Lambda_2$ are isometric if and only if there exists a linear isometry of $\Bbb R^2$ that sends $\Lambda_1$ to $\Lambda_2$.
Every diffeomorphism of $T^2$ $f$ is of the form $f=f_0\circ f_1$ for $f_0 \in \mathrm{Diff}_0(T^2)$ and $f_A$ is the left-multiplication by $A\in GL_2(\Bbb Z)$.
Yes, this is true for any diffeomorphism from $\Bbb R^2 / \Bbb Z^2$ to itself (so that the comment you added in edit does not make sense).
Finally, let me comment that I computed the area of your torus thanks to WolframAlpha, which seems to be close to 2.6. The usual torus has unit area. I might have made an error somewhere, but it seems to me that you torus isn't isometric to the standard flat torus.
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Your final sentence, slightly tweaked, gives the reason why this diffeomorphism $A$ does not exist: if yout take $\vec v_1 = \langle 1,0 \rangle$, then the geodesic induced by $\vec v_1$ does wrap around a circle in $T^2$, whereas the geodesic induced by the vector $A \vec v_1$ does not wrap around a circle. But a diffeomorphism must take a circle to a circle, so no such diffeomorphism as you have described can exist.
In general, consider a unit vector $\vec v_1$ making an angle $\theta = 2 \pi s$ from the positive $x$-axis (measured in radians). Consider also the geodesic in $T^2$ that is induced by $\vec v_1$. Then one of two possibilities holds:
Since a diffeomorphism cannot take a non-dense subset of $T^2$ to a dense subset of $T^2$, and since an "irrational rotation" diffeomorphism as you have described it would take a rational angle to an irrational angle, no such diffeomorphism exists.
I suspect what you are missing is that although a matrix $A \in GL_+(2,\mathbb R)$ does act on the universal covering space $\mathbb R^2$ of the torus $T^2$, a general matrix $A$ does not preserve the orbits of the lattice action of $\mathbb Z^2$ on $T^2$, and hence there is no induced diffeomorphism of $T^2$.
There is, nonetheless, a special subset of $GL_+(2,\mathbb R)$ whose elements do preserve orbits of the lattice action of $\mathbb Z^2$ on $T^2$ and which do induce diffeomorphisms of $T^2$: that subset is precisely the subgroup $GL_+(2,\mathbb Z)$.
(This is just to get the typo edit through.)