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Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\,\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually contained in $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\,\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\,\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

Note that the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

Note, one can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.