Suppose $M_1, dots M_k$ are smooth manifolds, then each of the projection maps $pi_i : M_1 times dots times...












3












$begingroup$



Suppose $M_1, dots M_k$ are smooth manifolds of dimensions $n_1, dots, n_k$ respectively, then each of the projection maps $pi_i : M_1 times dots times M_k to M_i$ is a smooth submersion




This was my attempted proof:



Proof: Let $M = M_1 times dots M_k$ and let $p = (p_1, dots, p_k) in M$ choose a smooth chart $(U, phi)$ containing $p$. Then $U = U_1 times dots times U_k$ and $phi = phi_1 times dots times phi_k$ where each $(U_j, phi_j)$ is a smooth chart in $M_j$. Also note that $phi = phi_1 times dots times phi_k$ means that $phi(p_1, dots, p_k) = (phi_1(p_1), dots, phi_k(p_k)) = left(phi_1 times dots times phi_kright)(p_1, dots, p_k)$.



Let's now compute the local coordinate representation of $widehat{pi_i}$, this is given by $widehat{pi_i} = phi_i circ pi_i circ phi^{-1} : phi[U] to phi_i[U_i]$. Observe that $phi^{-1} = phi_1^{-1} times dots times phi_k^{-1}$. Letting $(x^1, dots, x^{n_i})$ denote the component functions of $phi_i$ we have begin{align*}widehat{pi_i}(x^1, dots, x^{n_1 + dots + n_k}) &= phi_ibigg(pi_ibig(phi_1^{-1}(x^1, dots, x^{n_1 + dots + n_k}), dots, phi_k^{-1}(x^1, dots, x^{n_1 + dots + n_k})big)bigg) \
&=phi_ibigg( phi_i^{-1}(x^1, dots, x^{n_1 + dots + n_k})bigg)
\
&=(x^1, dots, x^{n_1 + dots + n_k})
end{align*}



Then $dwidehat{pi}_{phi(p)}$ is represented by the Jacobian matrix of $widehat{pi_i}$ at $phi(p)$ and computing the Jacobian matrix for $widehat{pi_i}$ we see that $$dwidehat{pi}_{phi(p)} = begin{bmatrix}
I & 0 \
0 & 0\
end{bmatrix}$$
with $n_i$ many $1$'s down the main diagonal. Thus since $d(pi_i)_p = d(widehat{pi_i})_{phi(p)}$ in the chart $(U, phi)$ we have $operatorname{rank}d(pi_i)_p = operatorname{rank}d(widehat{pi_i})_{phi(p)} = n_i = operatorname{dim}T_{p_i} M_i$ hence $d(pi_i)_p$ is surjective and thus $pi_i$ is a smooth submersion. $square$




Is my proof correct? If not where have I gone wrong?






proof-verification differential-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
    $endgroup$
    – Will M.
    Jan 24 at 5:13










  • $begingroup$
    You can try coordinate invariant approach as in here.
    $endgroup$
    – Sou
    Jan 24 at 5:59


















3












$begingroup$



Suppose $M_1, dots M_k$ are smooth manifolds of dimensions $n_1, dots, n_k$ respectively, then each of the projection maps $pi_i : M_1 times dots times M_k to M_i$ is a smooth submersion




This was my attempted proof:



Proof: Let $M = M_1 times dots M_k$ and let $p = (p_1, dots, p_k) in M$ choose a smooth chart $(U, phi)$ containing $p$. Then $U = U_1 times dots times U_k$ and $phi = phi_1 times dots times phi_k$ where each $(U_j, phi_j)$ is a smooth chart in $M_j$. Also note that $phi = phi_1 times dots times phi_k$ means that $phi(p_1, dots, p_k) = (phi_1(p_1), dots, phi_k(p_k)) = left(phi_1 times dots times phi_kright)(p_1, dots, p_k)$.



Let's now compute the local coordinate representation of $widehat{pi_i}$, this is given by $widehat{pi_i} = phi_i circ pi_i circ phi^{-1} : phi[U] to phi_i[U_i]$. Observe that $phi^{-1} = phi_1^{-1} times dots times phi_k^{-1}$. Letting $(x^1, dots, x^{n_i})$ denote the component functions of $phi_i$ we have begin{align*}widehat{pi_i}(x^1, dots, x^{n_1 + dots + n_k}) &= phi_ibigg(pi_ibig(phi_1^{-1}(x^1, dots, x^{n_1 + dots + n_k}), dots, phi_k^{-1}(x^1, dots, x^{n_1 + dots + n_k})big)bigg) \
&=phi_ibigg( phi_i^{-1}(x^1, dots, x^{n_1 + dots + n_k})bigg)
\
&=(x^1, dots, x^{n_1 + dots + n_k})
end{align*}



Then $dwidehat{pi}_{phi(p)}$ is represented by the Jacobian matrix of $widehat{pi_i}$ at $phi(p)$ and computing the Jacobian matrix for $widehat{pi_i}$ we see that $$dwidehat{pi}_{phi(p)} = begin{bmatrix}
I & 0 \
0 & 0\
end{bmatrix}$$
with $n_i$ many $1$'s down the main diagonal. Thus since $d(pi_i)_p = d(widehat{pi_i})_{phi(p)}$ in the chart $(U, phi)$ we have $operatorname{rank}d(pi_i)_p = operatorname{rank}d(widehat{pi_i})_{phi(p)} = n_i = operatorname{dim}T_{p_i} M_i$ hence $d(pi_i)_p$ is surjective and thus $pi_i$ is a smooth submersion. $square$




Is my proof correct? If not where have I gone wrong?






proof-verification differential-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
    $endgroup$
    – Will M.
    Jan 24 at 5:13










  • $begingroup$
    You can try coordinate invariant approach as in here.
    $endgroup$
    – Sou
    Jan 24 at 5:59
















3












3








3





$begingroup$



Suppose $M_1, dots M_k$ are smooth manifolds of dimensions $n_1, dots, n_k$ respectively, then each of the projection maps $pi_i : M_1 times dots times M_k to M_i$ is a smooth submersion




This was my attempted proof:



Proof: Let $M = M_1 times dots M_k$ and let $p = (p_1, dots, p_k) in M$ choose a smooth chart $(U, phi)$ containing $p$. Then $U = U_1 times dots times U_k$ and $phi = phi_1 times dots times phi_k$ where each $(U_j, phi_j)$ is a smooth chart in $M_j$. Also note that $phi = phi_1 times dots times phi_k$ means that $phi(p_1, dots, p_k) = (phi_1(p_1), dots, phi_k(p_k)) = left(phi_1 times dots times phi_kright)(p_1, dots, p_k)$.



Let's now compute the local coordinate representation of $widehat{pi_i}$, this is given by $widehat{pi_i} = phi_i circ pi_i circ phi^{-1} : phi[U] to phi_i[U_i]$. Observe that $phi^{-1} = phi_1^{-1} times dots times phi_k^{-1}$. Letting $(x^1, dots, x^{n_i})$ denote the component functions of $phi_i$ we have begin{align*}widehat{pi_i}(x^1, dots, x^{n_1 + dots + n_k}) &= phi_ibigg(pi_ibig(phi_1^{-1}(x^1, dots, x^{n_1 + dots + n_k}), dots, phi_k^{-1}(x^1, dots, x^{n_1 + dots + n_k})big)bigg) \
&=phi_ibigg( phi_i^{-1}(x^1, dots, x^{n_1 + dots + n_k})bigg)
\
&=(x^1, dots, x^{n_1 + dots + n_k})
end{align*}



Then $dwidehat{pi}_{phi(p)}$ is represented by the Jacobian matrix of $widehat{pi_i}$ at $phi(p)$ and computing the Jacobian matrix for $widehat{pi_i}$ we see that $$dwidehat{pi}_{phi(p)} = begin{bmatrix}
I & 0 \
0 & 0\
end{bmatrix}$$
with $n_i$ many $1$'s down the main diagonal. Thus since $d(pi_i)_p = d(widehat{pi_i})_{phi(p)}$ in the chart $(U, phi)$ we have $operatorname{rank}d(pi_i)_p = operatorname{rank}d(widehat{pi_i})_{phi(p)} = n_i = operatorname{dim}T_{p_i} M_i$ hence $d(pi_i)_p$ is surjective and thus $pi_i$ is a smooth submersion. $square$




Is my proof correct? If not where have I gone wrong?






proof-verification differential-geometry







share|cite|improve this question









$endgroup$





Suppose $M_1, dots M_k$ are smooth manifolds of dimensions $n_1, dots, n_k$ respectively, then each of the projection maps $pi_i : M_1 times dots times M_k to M_i$ is a smooth submersion




This was my attempted proof:



Proof: Let $M = M_1 times dots M_k$ and let $p = (p_1, dots, p_k) in M$ choose a smooth chart $(U, phi)$ containing $p$. Then $U = U_1 times dots times U_k$ and $phi = phi_1 times dots times phi_k$ where each $(U_j, phi_j)$ is a smooth chart in $M_j$. Also note that $phi = phi_1 times dots times phi_k$ means that $phi(p_1, dots, p_k) = (phi_1(p_1), dots, phi_k(p_k)) = left(phi_1 times dots times phi_kright)(p_1, dots, p_k)$.



Let's now compute the local coordinate representation of $widehat{pi_i}$, this is given by $widehat{pi_i} = phi_i circ pi_i circ phi^{-1} : phi[U] to phi_i[U_i]$. Observe that $phi^{-1} = phi_1^{-1} times dots times phi_k^{-1}$. Letting $(x^1, dots, x^{n_i})$ denote the component functions of $phi_i$ we have begin{align*}widehat{pi_i}(x^1, dots, x^{n_1 + dots + n_k}) &= phi_ibigg(pi_ibig(phi_1^{-1}(x^1, dots, x^{n_1 + dots + n_k}), dots, phi_k^{-1}(x^1, dots, x^{n_1 + dots + n_k})big)bigg) \
&=phi_ibigg( phi_i^{-1}(x^1, dots, x^{n_1 + dots + n_k})bigg)
\
&=(x^1, dots, x^{n_1 + dots + n_k})
end{align*}



Then $dwidehat{pi}_{phi(p)}$ is represented by the Jacobian matrix of $widehat{pi_i}$ at $phi(p)$ and computing the Jacobian matrix for $widehat{pi_i}$ we see that $$dwidehat{pi}_{phi(p)} = begin{bmatrix}
I & 0 \
0 & 0\
end{bmatrix}$$
with $n_i$ many $1$'s down the main diagonal. Thus since $d(pi_i)_p = d(widehat{pi_i})_{phi(p)}$ in the chart $(U, phi)$ we have $operatorname{rank}d(pi_i)_p = operatorname{rank}d(widehat{pi_i})_{phi(p)} = n_i = operatorname{dim}T_{p_i} M_i$ hence $d(pi_i)_p$ is surjective and thus $pi_i$ is a smooth submersion. $square$




Is my proof correct? If not where have I gone wrong?






proof-verification differential-geometry




proof-verification differential-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 5:07









PerturbativePerturbative

4,49121553




4,49121553












  • $begingroup$
    Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
    $endgroup$
    – Will M.
    Jan 24 at 5:13










  • $begingroup$
    You can try coordinate invariant approach as in here.
    $endgroup$
    – Sou
    Jan 24 at 5:59




















  • $begingroup$
    Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
    $endgroup$
    – Will M.
    Jan 24 at 5:13










  • $begingroup$
    You can try coordinate invariant approach as in here.
    $endgroup$
    – Sou
    Jan 24 at 5:59


















$begingroup$
Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
$endgroup$
– Will M.
Jan 24 at 5:13




$begingroup$
Your proof looked correct to me. I would simplify it as follows: "By induction suffices to consider $k = 2$ and then, by definition, there exists charts such that the local expression of $pi$ is simply $(x,y) mapsto y,$ which is clearly of class $mathscr{C}^infty$ and its derivative has full rank."
$endgroup$
– Will M.
Jan 24 at 5:13












$begingroup$
You can try coordinate invariant approach as in here.
$endgroup$
– Sou
Jan 24 at 5:59






$begingroup$
You can try coordinate invariant approach as in here.
$endgroup$
– Sou
Jan 24 at 5:59












1 Answer
1






active

oldest

votes


















1












$begingroup$

It is almost correct. The idea is right, but I would have written it like this:



Let $mathbf x_i in mathbf R^{n_i}$, so $mathbf x = (mathbf x_1,dots,mathbf x_k)$ is a point in $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$.



Then,
begin{align*}
widehatpi_i(mathbf x) = mathbf x_i,
end{align*}
or in matrix form as
$$
widehatpi_i =
begin{bmatrix}
0 & dotsb & I_i & dotsb & 0
end{bmatrix},
$$
where $I_i$ is the $n_itimes n_i$ identity matrix.
Since $dwidehat pi_i = widehat pi_i$, and $widehatpi_i$ clearly has rank $n_i$, $widehat pi_i$ is a submersion.



Note, in particular, what is wrong with your proof:




  • You have written that the coordinate form of $pi_i$ is actually the identity map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}tomathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$ by just not being careful.


  • The matrix for $widehat pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}to mathbf R^{n_i}$, so the matrix should have dimensions $n_itimes(n_1+dotsb+n_k)$.







share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085483%2fsuppose-m-1-dots-m-k-are-smooth-manifolds-then-each-of-the-projection-maps%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    It is almost correct. The idea is right, but I would have written it like this:



    Let $mathbf x_i in mathbf R^{n_i}$, so $mathbf x = (mathbf x_1,dots,mathbf x_k)$ is a point in $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$.



    Then,
    begin{align*}
    widehatpi_i(mathbf x) = mathbf x_i,
    end{align*}
    or in matrix form as
    $$
    widehatpi_i =
    begin{bmatrix}
    0 & dotsb & I_i & dotsb & 0
    end{bmatrix},
    $$
    where $I_i$ is the $n_itimes n_i$ identity matrix.
    Since $dwidehat pi_i = widehat pi_i$, and $widehatpi_i$ clearly has rank $n_i$, $widehat pi_i$ is a submersion.



    Note, in particular, what is wrong with your proof:




    • You have written that the coordinate form of $pi_i$ is actually the identity map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}tomathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$ by just not being careful.


    • The matrix for $widehat pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}to mathbf R^{n_i}$, so the matrix should have dimensions $n_itimes(n_1+dotsb+n_k)$.







    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      It is almost correct. The idea is right, but I would have written it like this:



      Let $mathbf x_i in mathbf R^{n_i}$, so $mathbf x = (mathbf x_1,dots,mathbf x_k)$ is a point in $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$.



      Then,
      begin{align*}
      widehatpi_i(mathbf x) = mathbf x_i,
      end{align*}
      or in matrix form as
      $$
      widehatpi_i =
      begin{bmatrix}
      0 & dotsb & I_i & dotsb & 0
      end{bmatrix},
      $$
      where $I_i$ is the $n_itimes n_i$ identity matrix.
      Since $dwidehat pi_i = widehat pi_i$, and $widehatpi_i$ clearly has rank $n_i$, $widehat pi_i$ is a submersion.



      Note, in particular, what is wrong with your proof:




      • You have written that the coordinate form of $pi_i$ is actually the identity map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}tomathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$ by just not being careful.


      • The matrix for $widehat pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}to mathbf R^{n_i}$, so the matrix should have dimensions $n_itimes(n_1+dotsb+n_k)$.







      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        It is almost correct. The idea is right, but I would have written it like this:



        Let $mathbf x_i in mathbf R^{n_i}$, so $mathbf x = (mathbf x_1,dots,mathbf x_k)$ is a point in $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$.



        Then,
        begin{align*}
        widehatpi_i(mathbf x) = mathbf x_i,
        end{align*}
        or in matrix form as
        $$
        widehatpi_i =
        begin{bmatrix}
        0 & dotsb & I_i & dotsb & 0
        end{bmatrix},
        $$
        where $I_i$ is the $n_itimes n_i$ identity matrix.
        Since $dwidehat pi_i = widehat pi_i$, and $widehatpi_i$ clearly has rank $n_i$, $widehat pi_i$ is a submersion.



        Note, in particular, what is wrong with your proof:




        • You have written that the coordinate form of $pi_i$ is actually the identity map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}tomathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$ by just not being careful.


        • The matrix for $widehat pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}to mathbf R^{n_i}$, so the matrix should have dimensions $n_itimes(n_1+dotsb+n_k)$.







        share|cite|improve this answer











        $endgroup$



        It is almost correct. The idea is right, but I would have written it like this:



        Let $mathbf x_i in mathbf R^{n_i}$, so $mathbf x = (mathbf x_1,dots,mathbf x_k)$ is a point in $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$.



        Then,
        begin{align*}
        widehatpi_i(mathbf x) = mathbf x_i,
        end{align*}
        or in matrix form as
        $$
        widehatpi_i =
        begin{bmatrix}
        0 & dotsb & I_i & dotsb & 0
        end{bmatrix},
        $$
        where $I_i$ is the $n_itimes n_i$ identity matrix.
        Since $dwidehat pi_i = widehat pi_i$, and $widehatpi_i$ clearly has rank $n_i$, $widehat pi_i$ is a submersion.



        Note, in particular, what is wrong with your proof:




        • You have written that the coordinate form of $pi_i$ is actually the identity map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}tomathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}$ by just not being careful.


        • The matrix for $widehat pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $mathbf R^{n_1}timesdotsbtimesmathbf R^{n_k}to mathbf R^{n_i}$, so the matrix should have dimensions $n_itimes(n_1+dotsb+n_k)$.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 24 at 5:22

























        answered Jan 24 at 5:16









        Alex OrtizAlex Ortiz

        10.9k21441




        10.9k21441






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085483%2fsuppose-m-1-dots-m-k-are-smooth-manifolds-then-each-of-the-projection-maps%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            Npm cannot find a required file even through it is in the searched directory

            Image
            up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more