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The canonical line bundle over $\mathbb{C}P^{n}$ is denoted by $\ell_{n}$. It is well known that: $$(\ell_{1}\otimes \ell_{1})\oplus1 \simeq\ell_{1}\oplus\ell_{1}$$ Both sides are isomorphic bundles as two topological vector bundles over $\mathbb{C}P^{1}$.The argument is explained in "K-theory and vector bundles" by Allen Hatcher. In fact the argument is based on the fact that the following two maps from $S^{1}$ to $GL_{2}(\mathbb{C})$ are homotopic maps $$\begin{pmatrix}z & 0\\ 0&z \end{pmatrix}\;\;\text{and}\begin{pmatrix}z^{2} & 0\\ 0&1 \end{pmatrix}$$

(A topological question):

How can we compare $(\ell_{n}\otimes \ell_{n})\oplus1 $ with $\ell_{n}\oplus\ell_{n}$. We know that their restriction to $\mathbb{C}P^{1}\subset \mathbb{C}P^{n}$ are isomorphic. So what is the difference element as an element of the relative K-theory $K(\mathbb{C}P^{n},\mathbb{C}P^{1})$?

(An algebra geometric question):

Is the isometry $$(\ell_{1}\otimes \ell_{1})\oplus1 \simeq\ell_{1}\oplus\ell_{1}$$ valid as an isomorphism of vector bundles in the context of algebraic geometry?

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    $\begingroup$ Compare Chern characters. $\endgroup$
    – Sasha
    Commented Sep 21, 2014 at 10:46
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    $\begingroup$ @Sasha could you please more explain: in $\mathbb{Z} [x]/x^{n+1}$ , we have ${(e^{x})}^{2}+1=2e^{x}$ so what is the contradiction? $\endgroup$ Commented Sep 21, 2014 at 15:47
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    $\begingroup$ No, this is false -- compare the coefficients of $x^2$. $\endgroup$
    – abx
    Commented Sep 21, 2014 at 17:16
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    $\begingroup$ As algebraic vector bundles, $(\ell_{1}\otimes \ell_{1})\oplus1$ and $\ell_{1}\oplus\ell_{1}$ -- or, in algebraic geometry language, $\mathcal{O}_{\mathbb{P}^1}(2)\oplus \mathcal{O}_{\mathbb{P}^1}$ and $\mathcal{O}_{\mathbb{P}^1}(1)^2$ -- are not isomorphic. Any algebraic vector bundle on $\mathbb{P}^1$ can be written $\mathcal{O}_{\mathbb{P}^1}(d_1)\oplus \ldots \oplus \mathcal{O}_{\mathbb{P}^1}(d_r)$ with $d_1\leq \ldots \leq d_r$, and this decomposition is unique. $\endgroup$
    – abx
    Commented Sep 21, 2014 at 17:23
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    $\begingroup$ $(e^x)^2+1 = e^{2x}+1 = 2 + 2x + 2x^2 + \frac43 x^3 + \dots$ while $2e^x = 2 + 2x + x^2 + \frac13 x^3 + \dots$ $\endgroup$
    – Sasha
    Commented Sep 21, 2014 at 17:40

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