Clarification on representing K-theory by vector bundles.












0












$begingroup$


In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:



'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
$$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$



The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?






topological-k-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:



    'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
    $$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$



    The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?






    topological-k-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:



      'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
      $$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$



      The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?






      topological-k-theory







      share|cite|improve this question









      $endgroup$




      In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:



      'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
      $$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$



      The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?






      topological-k-theory




      topological-k-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 1:15









      jojojojo

      6217




      6217






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.






          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%2f3092968%2fclarification-on-representing-k-theory-by-vector-bundles%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









            0












            $begingroup$

            To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.






                share|cite|improve this answer









                $endgroup$



                To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 1:27









                Tsemo AristideTsemo Aristide

                60.1k11446




                60.1k11446






























                    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%2f3092968%2fclarification-on-representing-k-theory-by-vector-bundles%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

                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                    Image
                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                    Read more

                    Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                    Image
                    up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                    Read more

                    A Topological Invariant for $pi_3(U(n))$

                    Image
                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                    Read more