# Tag Info

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Hint the determinant is continuous so the inverse image of the real line -0 by the determinant is an open subset so it is a submanifold.

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Before saying what I think are good introductions to mathematical gauge theory, I should say what I think gauge theory is. What gauge theory means to me is the application of certain PDEs, relevant in physics, to the topology and geometry of manifolds. I note here that one can be ignorant of the actual physics (as I am). Here is one example. Given a ...

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You're doing great so far. Let me use the name $\eta$ for the thing you call $\theta$, and then continue. You need to integrate $\eta$ over something that projects to the unit circle in the $xy$ plane. So let's take $$\gamma(t) = (\cos t, \sin t, f(\cos t, \sin t))$$ which parameterizes the set $\partial R_f$ nicely, and use it to compute a path ...

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The answer is yes for the following reason: $\omega_n$ is the Euler class mod 2 Now use e.g. Theorem 4.7 here which says $e(\nu_N)([N])$ counts number of intersections, or you argue that the Thom class of $\nu_N$ in $M$ is the Poincaré dual of $N$ (follows from this exercise). Hence, by pulling back to the cohomology of $N$ the result follows. You ...

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To show that the Jacobian really is of rank $1$ (i.e., not equal to the zero vector), I suggest you expand $\det A$ by minors along the first column, say: $$\det A = \sum_{i=1}^n (-1)^{i+1}a_{i,1}M_{i,1}.$$ Of course, $(-1)^{i+1}M_{i,1}$ is in fact an entry of the Jacobian. So the Jacobian can only be zero, if also $\det A=0$. But we are looking at the ...

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In other words, you are asking if there are "exotic" coequalisers in the category of manifolds. In fact, there are. Let $S^1 = \{ z \in \mathbb{C} : \left| z \right| = 1 \}$ be the circle, let $R = \mathbb{Z} \times S^1$, and let $d_0, d_1 : R \to S^1$ be defined as follows: $$d_0 (n, z) = \exp (i n) z$$ $$d_1 (n, z) = z$$ It is not hard to see that the ...

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Examining the Gauss map, we see that $\overline{g}$ is covered by the bundle map$$g \oplus g^\perp: \tau \oplus \nu \to \gamma_n \oplus \gamma^\perp$$and the differential$$D\overline{g}: \tau M \to \tau G_n(\mathbb{R}^{n+k}) \simeq \text{Hom}(\gamma^n, \gamma^\perp).$$This thus yields a fiber-linear morphism\sigma: \tau M \to \text{Hom}(\tau M, ...

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