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Pietro Majer
Pietro Majer
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To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and since $\|x\|=\|v\|=1$ we also get $x=\pm e_1$$x_1=\pm1$, and $v=\pm e_2$$v_2=\pm1$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and we get $x=\pm e_1$, $v=\pm e_2$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and since $\|x\|=\|v\|=1$ we also get $x_1=\pm1$, and $v_2=\pm1$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

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Pietro Majer
Pietro Majer
  • 60.6k
  • 4
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  • 269

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(q)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$$$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and we get $x=\pm e_1$, $v=\pm e_2$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(q)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and we get $x=\pm e_1$, $v=\pm e_2$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and we get $x=\pm e_1$, $v=\pm e_2$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

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Pietro Majer
Pietro Majer
  • 60.6k
  • 4
  • 122
  • 269

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(q)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and we get $x=\pm e_1$, $v=\pm e_2$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.