How can I show that the parametrized torus $T=\{(x,y,z)\in \mathbb{R}^3 : (\sqrt{x^2 +y^2}-a)^2 +z^2 =b^2 \}$ is a 2-dimensional smooth submanifold of $\mathbb{R}^3$ ? I was thinking of using the regular value theorem but I can't think of a smooth map that works. Any help would be greatly appreciated.
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For a calculus question  I need to parameterise the surface of the torus generated by rotating the circle given by $(x-b)^2+z^2=a^2$ around the $z$-axis (with $0&lt;a&lt;b$). I've had a go at this, and have come up with $$\vec{r}(\theta,\phi)=((a\cos\theta+b)\cos\phi, (a\cos\theta+b)\sin\phi, a\sin\theta), ~~ \theta,\phi\in [0,2\pi].$$  I haven't done anything like this for a while now, so I've kind of forgotten how it all goes. So I've tried plotting this on WolframAlpha to see if it's correct, but I can't get it to actually plot a surface defined in this way. I have a copy of Matlab, is there any way I could plot this surface on there to see if it's correct? Basically how can I check if I'm right or not? Failing that, could somebody please just point out if I'm wrong?
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I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!  The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$  The chain groups are  $$ C_0(T^2) = \{ k e^0 |&#160;k \in \mathbb{Z} \} = \mathbb{Z}$$  $$ C_1(T^2) = \{&#160;k_1 e^1_a + k_2 e^1_b |&#160;k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$  $$ C_2(T^2) = \mathbb{Z}$$  $$ C_k(T^2) = \{ 0 \} , k &gt; 2$$  Now the homology groups:  $$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$  where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)  $$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$  where $\ker \partial_1 =  \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.  $im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.  I don't want to use Hurewicz to do $H_1$.  $$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.  Thanks for your help.  Edit  I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them...
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I wish to show that the injection $k[y^2, y^3] \rightarrow k[y]$ is not flat.    I know of geometric ways to see this, but I wish to see explicitly $k[y^2, y^3]$-modules (or localizations thereof) $0 \rightarrow M' \rightarrow M$ which does not remain injective upon tensoring with $k[y]$. Is such a search futile?
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I was trying to prove that the horn torus and the spindle torus are not manifolds by definition(locally diffeomorphic to some Euclidean space.). I have no idea how to do this, but I attempted it in the following way:  I failed to show that a "slice" of manifold is a manifold itself. By a slice I mean, if the manifold $X\in \mathbf R^n$, then we set the coordinates $x_i=0$ for some $i$ where $0\leq i\leq n$. I feel this would work but I have no idea how to prove it.   Then I look at the cross section for the tori I mentioned(this is equivalent to taking a slice.). For the horn torus, you have two circles touching each other. And the fore spindle torus, you have two circles intersecting each other. Since a circle is locally diffeomorphic to $\mathbf R^1$, the two circles better have to be diffeomorphic to $R^1$. Otherwise we get the result we want.  I tried to proof that a neighbourhood around the point they touch(or intersect) cannot be locally diffeomorphic to $\mathbf R^1$. I tried to do it by contradiction. However I'm stuck on finding a contradiction...  Any thoughts?  P.S. Under the request of Sam Lisi, here are the definitions or horn and spindle tori:  A torus is the set of points in $\mathbf R^3$ at a distance $b$ from the circle of radius a in the $xy$ plane. This is like you put a circle with radius $b$ in the $yz$ plane centred at $(a,0,0)$. Then you make it orbit around the origin and you get a torus.  A horn torus is when $a=b$. If you take a cross-section, you'll find that it's two circles touching each other at one point. When $a&lt;b$, it's called a spindle torus. The cross-section would look like two circles intersecting with each other at two pionts.   There are pictures in this wikipedia article that might help you visualise the horn and spindle tori:
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I'm trying to get my head round the following calculation of the fundamental group of the torus, using Seifert Van-Kampen (I know it's easier to do this by considering covering spaces, but I'm trying to learn the Seifert Van-Kampen method).   Consider the torus $T$ as the unit square in $\mathbb R^2$ with opposite edges identified. Let $A$ be an open disc (say of radius $1/4$ about the origin), and $\overline{A}$ be it's closure. Let $B$ be complement of $A$, so that $A \cap B = S^1$. Then we have $T = A \cup B$ where $A$ and $B$ are a neighbourhood deformation retract pair, so we can use Van Kampen. It is clear that $\pi_1(A) = \{e \}$ and $\pi_1(A \cap B) = \mathbb Z$. So we need to find $\pi_1(B)$. Observe that $B$ deformation retracts onto the boundary of the square which, after identification, is homeomorphic to $S^1 \vee S^1$; in this way we see that $B$ and $S^1 \vee S^1$ are homotopy equivalent (inclusion in one direction, the retraction in another), and so $\pi(S^1 \vee S^1) = F_2$ (which can be calculated by Van Kampen, or by using the theory of covering spaces). So, by Van Kampen, $\pi_1(T) \cong F_2 *_{\mathbb Z} \{e\}$. This is the quotient of $F_2$ by the normal closure of the image of $l_* : \pi_1(A \cap B) \to \pi_1(B)$ (where $l_*$ is the homomorphism induced from the inclusion map $l : A \cap B \hookrightarrow B$). The following is causing me confusion: The generator of $\pi_1(A \cap B)$ is homotopic in $B$ to a loop of the form $u.v.\overline{u}.\overline{v}$, where $u$ and $v$ represent the two generators of $F_2$. So $\pi_1(T) \cong \langle a,b ; aba^{-1}b^{-1} \rangle$, the free abelian group on two generators.  Any explanation or geometric intuition would be really appreciated. Thanks
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I'm reading through G&amp;P's Differential Topology book, but I hit a wall at the end of section 4. There is a result stating     The set $X=\{A\in M_{m\times n}(\mathbb{R}):\mathrm{rk}(A)=r\}$ is a submanifold of $\mathbb{R}^{m\times n}$ with codimension $(m-r)(n-r)$.    There is a suggestion: Let $A\in M_{m\times n}(\mathbb{R})$ have form $$ A=\begin{pmatrix} B &amp; C \\ D &amp; E\end{pmatrix} $$ where $B$ is an invertible $r\times r$ matrix. Then right multiply by $$ \begin{pmatrix} I &amp; -BC^{-1} \\ 0 &amp; I \end{pmatrix} $$ and show $\mathrm{rk}(A)=r$ iff $E-DB^{-1}C=0$.  I multiplied out and got the matrix $$ M:=\begin{pmatrix} B &amp; 0 \\ D &amp; E-DB^{-1}C\end{pmatrix}. $$  Since I multiplied by a nonsingular matrix, I know that $\mathrm{rk}(A)=\mathrm{rk}(M)$. If $E-DB^{-1}C=0$, then  $$ M=\begin{pmatrix} B &amp; 0 \\ D &amp; 0 \end{pmatrix} $$ has rank $r$, so $A$ has rank $r$. For the converse, if $A$ has rank $r$, then $M$ has rank $r$, so by performing row operations, $M$ is row equivalent to a matrix of the form $$ \begin{pmatrix} I_r &amp; 0 \\ 0 &amp; 0 \end{pmatrix}. $$ This would imply $E-DB^{-1}C$ is row equivalent to $0$, and I think this implies $E-DB^{-1}C=0$.  My main concern is then, how does this approach imply $\mathrm{codim}(X)=(m-r)(n-r)$? Is there some special map I can apply the Preimage Theorem to?  Thank you.
46719
Let $U\subset\mathbb R^3$ be an open set, and $f:U\to \mathbb R$ be a smooth function. Suppose that the level set $S=f^{-1}(\{0\})$ is non-empty, and that at each $p\in S,$ the gradient $\overrightarrow \nabla f(p)$ is not the zero vector. Then $S$ is a smooth two-dimensional surface in $U$, and $p\mapsto \overrightarrow \eta(p)=\frac{1}{||\overrightarrow \nabla f(p)||}\overrightarrow \nabla f(p)$ defines a smooth unit-length normal vector field along $S$. At each $x\in U,$ write $H(f)_{(x)}$ for the $3\times 3$ Hessian matrix specified by $$(H(f)_{(x)})_{ij}=\frac{\partial^2f}{\partial x_i\partial x_j}(x).$$   Show that, at each $p\in S$, the second fundamental form $II_p: T_p(s)\times T_p(s)\to \mathbb R$ is the symmetric bilinear map    $$II_p(\overrightarrow v,\overrightarrow w)=\frac{-1}{||\overrightarrow \nabla f(p)||}\overrightarrow v\cdot H(f)_{(p)}\overrightarrow w,$$for all $\overrightarrow v ,\overrightarrow w \in T_p(s)$.       (Here, we view the tagent space $T_p(S)$ as the two-dimensional subspace $(span\{ {\overrightarrow \eta(p)}\})^{\bot}$ of $\mathbb R^3$.      Edit: Actually my question is why the second fundamental form under the usual definition can be written in this way.      Definition: The quadratic form $II_p$, defined in $T_p(S)$ by $II_p(v)=-&lt;d  N_p(v),v&gt;$ is called the second fundamental form of $S$ at $p$, where $dN_p:T_p(S)\to T_p(S)$ is the differential of the Gauss map.   Hopefully, I express this problem explicitly. I was just wondering how to prove this statement.   I took a diffrential geometry class last semester, and when I organized my notes this morning, I found this statement, but there was no proof...  Looking forward to an understandable explaination. Thanks in advance.     Edit 2:Furthermore, show that, at each point $p\in S$, the expression   $$\phi_p(z)=det\pmatrix{-H(f)_{(p)}-zI_{3\times 3} &amp; \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)&amp; 0}$$   (the underlying matrix here is $4\times 4$) defines a second-degree polynomial whose roots $\lambda_1$ and $\lambda_2$ are $||\overrightarrow \nabla f(p)||k_1$ and $||\overrightarrow \nabla f(p)||k_2$, where $k_1$ and $k_2$ are the principal curvatures of $S$ at $p$.      Also, if a non-zero vector $\pmatrix {\overrightarrow v \\c}$ lies in the kernel of the $4\times 4$ matrix $$\pmatrix{-H(f)_{(p)}-\lambda_jI_{3\times 3} &amp; \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)&amp; 0},$$   then $\vec v$ is a non-zero element of $T_p(S)$ and lies in the "principal direction" corresponding to $K_j$.
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In class today our teacher mentioned that one can use the implicit function theorem to show that $O(n) \subseteq \mathbb{R}^{n^2}$ is a submanifold...that is, map $A \mapsto A^* A$, and set it equal to the identity matrix. There should be $n(n+1)/2$ equations in $n^2$ variables ($A^*A$ is always symmetric, so we only need $n(n+1)/2$ equations). When I asked how to show that the jacobian matrix of this transformation had full row rank (the condition of the implicit function theorem), my teacher said "take the derivative near the identity, and you see that the derivative is just a matrix plus its transpose ."  I wasn't sure what he meant. Can someone clarify?
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In Hartshorne section 1.1 he gives a problem (ex 1.11) which says that,   Let $Y \subset \mathbb A^3$ be the curve given parametrically by $x=t^3, y=t^4, z=t^5$. Show that $I(Y)\subset k[x,y,z]=A$ is a prime ideal of height 2 which cannot be generated by two elements.  Firstly, my question is that how can I check whether $Y$ is Zariski closed or not. Is it always the case that, a subset given by $x=f(t), y=g(t), z=h(t); f,g,h \in k[t]$ is always Zariski closed ?  My attempt was to prove that $A/I(Y) \cong k[t^3,t^4,t^5]$ to show $I(Y)$ is prime. And then I wanted to show the map $ \mathbb A^1 \rightarrow \mathbb A^3$ sending $t \rightarrow (t^3,t^4,t^5)$ is a topological embedding. I am stuck in proving that the map is continuous.   After proving this I could have said $Y$ (having proved that it is Zariski closed) is a variety of dimension 1, hence $I(Y)$ has height 2. I don’t know how to determine the least number of generators. Any help from basic algebraic results would be helpful.  I am a beginner in algebraic geometry so I don’t know many things, please keep it as basic as possible.
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I need some help on the following problem:      Given that a curve $\mathbf r:I\to \Bbb R ^3$ has constant curvature $k(s)=k$, for all $s$, and constant torsion $\tau(s)=\tau$, for all $s$. Find the curve $\mathbf r$.    I only know that, according to the fundamental theorem, this curve exists and is unique. But, how practically find the parametric equation of the curve?   Thanks.
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I'm trying to show the following:  Let $f:X\to Y$ be a proper holomorphic map between connected, non-empty Riemann Surfaces. Show that a map $g:Y\to\mathbb{C}$ is  holomorphic if and only if  its composition with f is holomorphic.  So far I know that $f$ is surjective, since it's proper and holomorphic. My next step was to look at the charts, but it's there, I get stuck.
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I am using Do Carmo's Riemannian Geometry, and struggling to solve a problem.  The problem is:  Show that the mapping $F:\mathbb{R}^2\to\mathbb{R}^4$ given by  $$F(x,y)=((r\cos y+a)\cos x,(r\cos y+a)\sin x,r\sin y\cos\frac{x}{2},r\sin y\sin\frac{x}{2}))$$  induces an embedding of the Klein bottle into $\mathbb{R}^4$.  I know that the Klein bottle is defined as a quotient manifold $T^2/G$, where $T^2$ is the 2-torus, $G$ is a group of diffeomorphisms of $T^2$ formed by $\{A,Id\}$, and $A$ is the antipodal map, i.e. $A(p)=-p$.  Besides, I know that $T^2=S^1\times S^1$, where $S^1$ is the unit circle.  Also, I found that $F$ is injective.  However, I don't have a clue about how to attack this problem. I tried googling, but found nothing.  Any help will be appreciated.
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In the definition of a parametrised surface $S$, for every point in the surface, $p \in W \subseteq S$, where $W$ is open, there exists a coordinate chart or patch , $F :U\to \mathbb{R}^n$ that maps to $p$ from an open subset $U \in \mathbb{R}^n$  Is that right? If anyone knows of a more general definition, I'm willing to learn. It sounds a lot like a manifold, which I'm not entirely familiar with.  In this definition, the number of surface patches in the atlas is not stipulated. Given a parametrisable surface, is a finite number of charts sufficient to describe the surface? Can we find a surface that requires infinitely many patches to fully chart? If so, in which $\mathbb{R}^n$ does the first such surface occur?   In which dimensions is it always possible to find a finite number of patches for any given surface?  EDIT: Added requirement that such a surface (manifold) be connected. One made from infinitely many disconnected subsets would have to be charted infinitely.
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Compute the fundamental group of the space obtained from two tori $S^1✕S^1$ by identifying a circle $S^1✕${$x_0$} in one torus with the corresponding circle $S^1✕${$x_0$} in the other.  Using van Kampen's theorem, I can fairly quickly show that if $T_1$ is one torus and $T_2$ is the other, then the group has to be isomorphic to $((π_1(T_1) ∗ π_1(T_2)))/N ≅ (\mathbb{Z} ✕ \mathbb{Z})∗ (\mathbb{Z} ✕ \mathbb{Z})/N$, where N is generated by elements of the form $i_{12}(w)i_{21}(w)^{-1}$, w is in $\pi_1(S^1✕${$x_0$}$) ≅ \mathbb{Z}$ and $i_{12}$, $i_{21}$ are the maps from $\pi_1(S^1✕${$x_0$}$)$ to $π_1(T_1)$, $π_1(T_2)$ respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like. My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like $i_{12}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_1$ and $i_{21}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_2$, which says to me that $i_{12}(w)i_{21}(w)^{-1} $&quot;$=$&quot;$ (n, 0) + (-n, 0) = 0$. The notation here isn't right, since $(n, 0)$ in $T_1$ is not the same as $(n, 0)$ in $T_2$ and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in $S^1✕${$x_0$} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in $T_1$ and d, e are the generators of the circles in $T_2$, and c is the constant path then the identification of the circle should make, say, $a = d$, so for example $(a, b)*(d, e)$ $ = (a^2,b)*(c, e)$ $ = (c, b)*(d^2,e)$. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense. My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the &quot;2-torus&quot;, 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus $S^1✕S^1✕S^1$ or something completely different.
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I am reading Lee's book on Differential geometry. In Chapter 4, Lee has this exercise.  Let $M$, $N$ be smooth manifolds, and let $f:M\to N$ be a smooth map. Define $F : M\to M \times N$ by $F(x) = (x, f(x) )$. Show that for every smooth vector field $V$ of $M$ , there is a smooth vector field on $W$ on $M \times N$ that is $F$-related to $V$.  I can understand that we must have $W_{(x,f(x))}=V_x\oplus Df(x)V_p$ for all $x\in M.$ The set $\{(x,f(x)):x\in M\}$ is a closed set in $M\times N.$ Therefore, if we can show that for all $(p,f(p))\in M\times N$ there is a neighbourhood $U_p$ and a smooth vector field on $U_p$ extending $V_x\oplus Df(x)V_x$ we are done by partition of unity. But I cannot show that. Can someone help me out?
20170
I am looking for any help, hints, or suggestions in how to go about this problem from a previous qualifying exam.  We are given a smooth mapping $f: U \rightarrow \mathbb{R}^{n}$ whose differential $df_{p}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is of rank $n$ and we have that $\|f(x)-x\|&lt;\frac{1}{2}$. Here $U \subset \mathbb{R}^{n}$ is an open subset which contains the closed unit ball $B=B_{1}(0)$. The problem asks us to show that $\|f\|^{2}$ attains a minimum on the interior of  $B$.   As of right now, the only thing that comes to mind is the inverse function theorem, that there is an open subset $V \subset U$ where $f$ is an diffeomorphism on.
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I have the following problem: Let $(M, \omega)$ be a symplectic manifold. How can I show $$\omega^n=\underbrace{\omega\wedge \ldots\wedge \omega}_{n-times},$$ satisfies $\omega^n(p)\neq 0$ for all $p\in M$. I believe that is not too dificult but I'm not used with exterior product.. Any help will be valuable...  This is important because the nondegeneracy condition on $ω$ is equivalent to the condition that M has an even dimension $2n$ and the top wedge product $\omega^n$ is nowhere vanishing on M, i.e., $ω^n$ is a volume form. In particular, $M$ must be orientable and is canonically oriented by $\omega^n$.