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As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempted a modification to make the map look a little more like Bott periodicity.

I think that the answer is a qualified no. It's an interesting map, but as I see On the face of it, it Eric's map does not carry the same information as the Bott map. The homotopy information Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in the domainmind, the targetas I understand now, and the map all change is a lot. Of course, with enough new ideas you could try to change the map result of Snaith that constructs a spectrum equivalent to make it the Bott mapspectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Snaith's model has been called "Snaith periodicity", but I think the existing arguments that such an effort would be it is the same are a use and not a proof of Bott periodicity. (In that sense, Snaith's model is stone soup.

First, I think although that the Bott metaphor is not really fair to his good paper.)

For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx casesFor complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map$$\Sigma X \to U(2n).$$

At first glance, Eric's twisted suspension only is very different. It exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. If course $X$ \mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a line bundle totally different homotopy structure from $L$, then it has a map BU(\infty)$. Moreover, twisted suspensions aren't adjoint to $\mathbb{C}P^\infty$. Eric thus constructs a mapordinary delooping. Instead, the space of maps $$\Sigma^L \Sigma^L X \to \mathbb{C}P^\infty,$$which means that the twisted suspension also has Y$ is adjoint to sections of a line bundle $L$. You can check that over $\Sigma^L X$ with fiber $\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed.Moreover

However, the purpose of the suspension maps in spectra Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott periodicity element $\beta$. (I think that the "+" subscript just denotes choosing a base point.) You can see what is to go coming just from a map the rational homotopy groups of $\Sigma X \Sigma^\infty \to Y$ to its adjoint map mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $X \K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to \mathcal{L}Y$. In this casethe theorem that stable homotopy groups of spheres are finite.) Moreover, since it is a in stable, rational homotopy, twisted and untwisted suspension , you instead get a bundle over become the same. So Snaith's model is built from the fact that the homology of $X$ with fiber \mathbb{C}P^\infty$ equals the homotopy of $\mathcal{L}^2\mathbb{C}P^\infty$. Again, this could be interestingBU(\infty)$. Moreover, but it there is not Bott periodicity.

Finally Eric's step 3 formally desuspends both sides to getan important determinant map$$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty $\det:BU(\infty) \to BU(1) = \Sigma^{-2}\mathbb{C}P^\infty.$$You can't desuspend unless you first suspend infinitely many times mathbb{C}P^\infty$$that takes the direct sum operation for bundles to get a spectrumtensor multiplication of line bundles. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on Snaith makes a yet different trackmoral inverse to this map (and not just in rational homology).Without changes

Still, this searching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a really complicated spectrum reminiscent purely algebraic proof of the sphere spectrumfundamental theorem of algebra. If $\mathbb{C}P^\infty$ The fundamental theorem of algebra is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ not a purely algebraic statement! It is too complicated to resemble itan analytic theorem with an algebraic conclusion, since the complex numbers are defined analytically. Edit: Eric points out that Snaith actually established The best you can do is a way to modify mostly algebraic proof, using some minimal analytic information such as that $\Sigma_+^{\infty}\mathbb{C}P^\infty$ (\mathbb{R}$ is real-closed using the plus sign has to do with choosing base pointsintermediate value theorem. Likewise, according to one reference) to make it $BU(\infty)$. What Bott periodicity is not clear to me a purely homotopy-theoretic theorem; it is whether this result a Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof of Bott periodicity or a that carefully uses as little Lie theory as possible. The proof by Bruno Harris fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use of Bott periodicityabout the unitary groups.

(The answer is significantly revised now that I know more about Snaith's result.)
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Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempted modification to make the map look a little more like Bott periodicity.

I think that the answer is no. It's an interesting map, but as I see it, it does not carry the same information as the Bott map. The homotopy information in the domain, the target, and the map all change a lot. Of course, with enough new ideas you could try to change the map to make it the Bott map, but I think that such an effort would be stone soup.

First, I think that the Bott map as Bott constructed it in the Annals is beautiful. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx cases, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map $$\Sigma X \to U(2n).$$ Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group.

(If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)

Eric's twisted suspension only exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. If $X$ is a space with a line bundle $L$, then it has a map to $\mathbb{C}P^\infty$. Eric thus constructs a map $$\Sigma^L X \to \mathbb{C}P^\infty,$$ which means that the twisted suspension also has a line bundle $L$. You can check that $\Sigma^L X$ depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. Moreover, the purpose of the suspension maps in spectra and Bott periodicity is to go from a map $\Sigma X \to Y$ to its adjoint map $X \to \mathcal{L}Y$. In this case, since it is a twisted suspension, you instead get a bundle over $X$ with fiber $\mathcal{L}^2\mathbb{C}P^\infty$. Again, this could be interesting, but it is not Bott periodicity.

Finally Eric's step 3 formally desuspends both sides to get $$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty \to \Sigma^{-2}\mathbb{C}P^\infty.$$ You can't desuspend unless you first suspend infinitely many times to get a spectrum. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on a yet different track. This Without changes, this is a really complicated spectrum reminiscent of the sphere spectrum. If $\mathbb{C}P^\infty$ is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ is too complicated to resemble it. Edit: Eric points out that Snaith actually established a way to modify $\Sigma_+^{\infty}\mathbb{C}P^\infty$ (the plus sign has to do with choosing base points, according to one reference) to make it $BU(\infty)$. What is not clear to me is whether this result is a proof of Bott periodicity or a use of Bott periodicity.
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Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempted modification to make the map look a little more like Bott periodicity.

I think that the answer is no. It's an interesting map, but as I see it, it does not carry the same information as the Bott map. The homotopy information in the domain, the target, and the map all change a lot. Of course, with enough new ideas you could try to change the map to make it the Bott map, but I think that such an effort would be stone soup.

First, I think that the Bott map as Bott constructed it in the Annals is beautiful. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx cases, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map $$\Sigma X \to U(2n).$$ Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group.

(If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)

Eric's twisted suspension only exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. If $X$ is a space with a line bundle $L$, then it has a map to $\mathbb{C}P^\infty$. Eric thus constructs a map $$\Sigma^L X \to \mathbb{C}P^\infty,$$ which means that the twisted suspension also has a line bundle $L$. You can check that $\Sigma^L X$ depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. Moreover, the purpose of the suspension maps in spectra and Bott periodicity is to go from a map $\Sigma X \to Y$ to its adjoint map $X \to \mathcal{L}Y$. In this case, since it is a twisted suspension, you instead get a bundle over $X$ with fiber $\mathcal{L}^2\mathbb{C}P^\infty$. Again, this could be interesting, but it is not Bott periodicity.

Finally Eric's step 3 formally desuspends both sides to get $$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty \to \Sigma^{-2}\mathbb{C}P^\infty.$$ You can't desuspend unless you first suspend infinitely many times to get a spectrum. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on a yet different track. This is a really complicated spectrum reminiscent of the sphere spectrum. If $\mathbb{C}P^\infty$ is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ is too complicated to resemble it.