Question about Chern Character in Hatcher's book The 2019 Stack Overflow Developer Survey Results Are InQuestion involving the Chern character from the book “Fibre Bundles”Chern Character Isomorphism for non-CW complexeschern character of wedge product of bundlesQuestion about Hatcher's book CW complexAbout homotopy fiber at Hatcher's bookUniqueness of the Chern characterHatcher's KTVB - Proposed Isomorphismchern character domain and codomainQuestion about proof of Kunneth theorem in Hatcher's bookquestion about a notation in Atiyah's book K-Theory

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

What is this business jet?

Can you cast a spell on someone in the Ethereal Plane, if you are on the Material Plane and have the True Seeing spell active?

Is it correct to say the Neural Networks are an alternative way of performing Maximum Likelihood Estimation? if not, why?

How do I free up internal storage if I don't have any apps downloaded?

What do these terms in Caesar's Gallic wars mean?

Deal with toxic manager when you can't quit

Why does the nucleus not repel itself?

Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?

Did any laptop computers have a built-in 5 1/4 inch floppy drive?

Why couldn't they take pictures of a closer black hole?

Are spiders unable to hurt humans, especially very small spiders?

Loose spokes after only a few rides

How to type a long/em dash `—`

Is Cinnamon a desktop environment or a window manager? (Or both?)

Can withdrawing asylum be illegal?

Correct punctuation for showing a character's confusion

Match Roman Numerals

How can I add encounters in the Lost Mine of Phandelver campaign without giving PCs too much XP?

What information about me do stores get via my credit card?

Star Trek - X-shaped Item on Regula/Orbital Office Starbases



Question about Chern Character in Hatcher's book



The 2019 Stack Overflow Developer Survey Results Are InQuestion involving the Chern character from the book “Fibre Bundles”Chern Character Isomorphism for non-CW complexeschern character of wedge product of bundlesQuestion about Hatcher's book CW complexAbout homotopy fiber at Hatcher's bookUniqueness of the Chern characterHatcher's KTVB - Proposed Isomorphismchern character domain and codomainQuestion about proof of Kunneth theorem in Hatcher's bookquestion about a notation in Atiyah's book K-Theory










2












$begingroup$


I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



enter image description here



We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



My question is why holds



$$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



The author refers to a recursion formula from page 63:



$s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



where $sigma_k$ are the $k$-th symmetric polynomials.



What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



Thank you.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



    enter image description here



    We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



    Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



    My question is why holds



    $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



    The author refers to a recursion formula from page 63:



    $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



    where $sigma_k$ are the $k$-th symmetric polynomials.



    What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



    Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



    Thank you.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



      enter image description here



      We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



      Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



      My question is why holds



      $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



      The author refers to a recursion formula from page 63:



      $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



      where $sigma_k$ are the $k$-th symmetric polynomials.



      What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



      Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



      Thank you.










      share|cite|improve this question











      $endgroup$




      I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



      enter image description here



      We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



      Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



      My question is why holds



      $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



      The author refers to a recursion formula from page 63:



      $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



      where $sigma_k$ are the $k$-th symmetric polynomials.



      What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



      Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



      Thank you.







      algebraic-topology vector-bundles topological-k-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 5:02









      Andrews

      1,2812423




      1,2812423










      asked Mar 24 at 4:48









      KarlPeterKarlPeter

      6771416




      6771416




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            I always find it helps to think about these things in terms of the Splitting Principle:




            For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




            If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



            where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



            The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



            $$ s_n = (-1)^n n sigma_n $$






            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%2f3160110%2fquestion-about-chern-character-in-hatchers-book%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






                  share|cite|improve this answer









                  $endgroup$



                  I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 24 at 5:36









                  Eric WofseyEric Wofsey

                  193k14220352




                  193k14220352





















                      1












                      $begingroup$

                      I always find it helps to think about these things in terms of the Splitting Principle:




                      For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                      If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                      where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                      The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                      $$ s_n = (-1)^n n sigma_n $$






                      share|cite|improve this answer











                      $endgroup$

















                        1












                        $begingroup$

                        I always find it helps to think about these things in terms of the Splitting Principle:




                        For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                        If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                        where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                        The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                        $$ s_n = (-1)^n n sigma_n $$






                        share|cite|improve this answer











                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          I always find it helps to think about these things in terms of the Splitting Principle:




                          For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                          If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                          where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                          The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                          $$ s_n = (-1)^n n sigma_n $$






                          share|cite|improve this answer











                          $endgroup$



                          I always find it helps to think about these things in terms of the Splitting Principle:




                          For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                          If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                          where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                          The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                          $$ s_n = (-1)^n n sigma_n $$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 24 at 6:01

























                          answered Mar 24 at 5:39









                          WilliamWilliam

                          3,2311228




                          3,2311228



























                              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%2f3160110%2fquestion-about-chern-character-in-hatchers-book%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

                              How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

                              random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

                              Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye