Sphere eversions Regular homotopy classes of immersions $S^n\to \mathbb{R}^{n+1}$ are classified by the group $\pi_n(SO(n+1))$. In particular, it follows from the fact that $\pi_2(SO(3))=0$ that an immersed 2-sphere in $\mathbb{R}^3$ can be turned "inside out," (called a sphere eversion). 
For $n=4$, we have $\pi_4(SO(5))=\mathbb{Z}/2\mathbb{Z}$, so there are two regular homotopy classes of immersions $S^4\to \mathbb{R}^5$.
What do these two classes correspond to? My guess is that they correspond to the two different orientations of $S^4$. So up to orientation, there is a unique regular homotopy class of immersions $S^4\to \mathbb{R}^4$. Is this correct?
 A: The short answer is "yes" but instead I'll explain the origin of the obstruction class in $\pi_n SO(n+1)$ that determines if a given immersion $S^n \to \Bbb R^{n+1}$ can be isotoped to the standard embedding.
Suppose $M^m, N^n$ are manifolds, $M$ is closed, with $m < n$. The space of $1$-jets of maps $M \to N$, denoted as, $J^1(M, N)$ encodes "all local first order differential restrictions" that can be imposed on a map $M \to N$. Specifically, elements of $J^1(M, N)$ consist of equivalence classes of triples or "1-jets" $(x, g(x), dg_x)$ for a point $x \in M$ and a smooth germ $g \in C^\infty_x(M, N)$ defined near $x$.
There is a vector bundle $p : J^1(M, N) \to M \times N$ given by $(x, g(x), dg_x) \mapsto (x, g(x))$ with fiber over the point $(x, y)$ being $\text{Hom}(T_x M, T_y N)$. Denote $\pi : J^1(M, N) \to M$ to be the fibration $\pi_1 \circ p$. Notice that any map $f : M \to N$ gives rise to a section $j^1 f : M \to J^1(M, N)$ of $\pi$, defined by $j^1 f(x) = (x, f(x), df_x)$. Such sections are known as "holonomic", and are a specific class of sections of $\pi$. If $s \in \Gamma(M, J^1(M, N))$, in general $\text{bs}(s):= p \circ s : M \to N$ (the "base" of $s$) and the third component of $s$, which is choice of an element of $\text{Hom}(T_x M, T_{\text{bs}(s)(x)} N)$ for every $x \in M$ will have nothing to do with each other. If $\text{bs}(s)$ is imagined as a graph of a function in $M \times N$, $s$ is merely that graph with a non-vertical distribution along that graph given by graph of the linear map $T_x M \to T_{\text{bs}(s)(x)} N$ in $T_{(x, \text{bs}(s)(x))} M \times N$. If $s$ is holonomic, then this distribution is integrable and is tangential to the graph of $\text{bs}(s)$. Solutions to first-order partial differential equations imposed on maps $M \to N$ are same as holonomic sections with image in closed submanifolds of $J^1(M, N)$
Call $\mathscr{R}_\text{imm} \subset J^1(M, N)$ to be the subset consisting of $1$-jets of germs of immersions. Equivalently, elements of $\mathscr{R}_\text{imm}$ consist of triples of the form elaborated above with third component injective. This is a subbundle of $p$ with fibers over $(x, y) \in M \times N$ being $\text{Emb}(T_x M, T_y M)$. $\mathscr{R}_\text{imm}$ is known to be the "differential relation" of immersions (this is an open submanifolds of $J^1(M, N)$, so it's not quite a partial differential equation in the sense described above, but perhaps akin to a partial differential inequality). Denote by $\text{Sec}(\mathscr{R}_\text{imm})$ to be the space of sections $s : M \to J^1(M, N)$ of $\pi$ such that $s(M) \subset \mathscr{R}_\text{imm}$, i.e, $s$ takes values on $1$-jets of germs of immersions. Intuitively $\text{Sec}(\mathscr{R}_\text{imm})$ consists of smooth functions $M \to N$ "imposed with the differential restriction $\mathscr{R}_\text{imm}$". Let $\text{Hol}(\mathscr{R}_\text{imm}) \subset \text{Sec}(\mathscr{R}_\text{imm})$ consist of the subspace of such sections which are moreover integrable (or holonomic), i.e., appear as 1-jet prolongation of some global element of $C^\infty(M, N)$. Then we have the following fantastic theorem (also known as the $h$-principle for $\mathscr{R}_\text{imm}$)
Theorem: The inclusion $\text{Hol}(\mathscr{R}_\text{imm}) \hookrightarrow \text{Sec}(\mathscr{R}_\text{imm})$ is a weak homotopy equivalence. Moreover, the "weak deformation retract is $C^0$-small", in the sense that any $S^n$-parameterized family of sections $s: S^n \to \text{Sec}(\mathscr{R}_\text{imm})$ can be realized as a $C^0$-limit of a sequence $s_k: S^n \to \text{Hol}(\mathscr{R}_\text{imm})$ of $S^n$-parameterized family of holonomic sections.
(The reason it is so fantastic to me is that it essentially states non-integrable objects can be approximated by integrable objects. But this is blatant garbage, as integrability is a closed condition. The resolution of this paradox is ultimately that the approximation is of $C^0$-regularity)
This is Gromov's formulation of the Smale-Hirsch theorem, and can be reformulated as follows. A section $s : M \to \mathscr{R}_\text{imm}$ of $\pi$ can be envisioned as a map of pairs $(F, f) : (TM, M) \to (TN, N)$ where $f: M \to N $ is a smooth map, $F: TM \to TN$ a fiberwise embedding of the total spaces of the tangent bundles, and $M$, $N$ are realized as zero sections of $TM$, $TN$ respectively. Here $f = \text{bs}(s)$ is the base of section, and $F$ is fiberwise simply the third component of $s$. Demanding that $s$ is holonomic, i.e., $s = j^1 f$ is equivalent to claiming $F = df$, hence in particular that $f$ is an immersion. Therefore we recover the original Smale-Hirsch theorem:
Theorem: The map $\text{Imm}(M, N) \to \text{Emb}(TM, TN)$ given by $f \mapsto (df, f)$ is a weak homotopy equivalence.
In particular, this gives an isomorphism $\pi_0 \text{Imm}(M, N) \to \pi_0 \text{Emd}(TM, TN)$. To prove sphere eversion, take $M = S^n$ and $N = \Bbb R^{n+1}$. Then $\pi_0 \text{Imm}(S^n, \Bbb{R}^{n+1})$ classifies, precisely, immersions of the sphere $S^n \to \Bbb R^{n+1}$ upto isotopy-through-immersions. Notice that given a bundle-embedding $F : TS^n \to T\Bbb R^{n+1}$, there is a natural extension $\widetilde{F} : S^n \times \Bbb R^{n+1} \to T\Bbb R^{n+1}$ to the trivial $\Bbb R^{n+1}$-bundle on $S^n$ by extending over the orthogonal complement, using stable triviality of $TS^n$. (This is the so-called "microextension trick", which is essential to the proof of Smale-Hirsch theorem: essentially one wants extra dimension to apply the corrugations, which are explained in more concrete terms in the famous Geometry Center video that are crucial to the wild $C^0$-approximation of the non-holonomic sections by holonomic ones, which microextension provides). This gives rise to a "classifying map" $S^n \to \text{GL}(n+1)$, but there is an extra care required. $F$ might be orientation-reversing on the image, in which case extending to $\widetilde{F}$ by identity on the orthogonal complement is not the correct approach: that gives rise to an extra obstruction coming from the classifying map $S^n \to \text{GL}(n+1)$ having image inside $\text{GL}^+(n+1)$ or $\text{GL}^-(n+1)$, which was not intrinsic to $F$ as orientation-reversing on the image is irrelevant to $F$ being fiberwise homotopic to a different bundle-embedding which might be orientation-preserving on the image. Therefore by microextending so that $\widetilde{F}$ is orientation-preserving fiberwise (which can always be done by demanding orientation preserval or reversal on the normal component), one obtains a classifying map $S^n \to \text{GL}^+(n+1) \simeq SO(n+1)$, homotopy classes of which classify fiberwise homotopy classes of bundle embeddings $TS^n \to T\Bbb R^{n+1}$.
This is the required isomorphism $\pi_0 \text{Imm}(S^n, \Bbb R^{n+1}) \to \pi_n SO(n+1)$ $\blacksquare$
Consider the regular homotopy class $[-\text{id}]$ of the antipodal embedding and let $\alpha$ be the corresponding element of $\pi_n SO(n+1)$. The microextended immersion in this case is simply $f : \nu(S^n) \to \Bbb R^{n+1}$ given by $f(x, \mathbf{n}) = (-x, (-1)^{n+1} \mathbf{n})$ where the factor of $(-1)^{n+1}$ comes from the sign consideration mentioned earlier to ensure $f$ is orientation-preserving, given that $-\text{id}$ is orientation-reversing iff $n$ is even. Let $\mathcal{G} : S^n \to \text{Grass}(n, n+1)$ be the tangential Gauss map of the standard $S^n \subset \Bbb R^{n+1}$, and denote by $M$ to be the matrix $$M = \begin{pmatrix}I_{n \times n} & \mathsf{O}_{n \times 1} \\ \mathsf{O}_{1 \times n} & (-1)^{n+1} \end{pmatrix}$$ Then $\alpha$ is simply represented by the map $\mathcal{A}: S^n \to \text{GL}^+(n+1)$ defined as $\mathcal{A}_x = T_x^{-1} \mathcal{G} T_x$ where $T_x : (\Bbb R^{n+1}, \mathcal{G}_x) \to (\Bbb R^{n+1}, \Bbb R^n \times 0)$ is establishing a standardization of $\Bbb R^{n+1}$ realized as $\mathcal{G}_x \oplus \mathcal{G}_x^\perp$
$\mathcal{A}$ is not nullhomotopic as it's the clutching function of the tangent bundle $TS^{n+1}$. To see this, consider a standard frame $E^+ = (e^+_1, \cdots, e^+_{n+1})$ at the north pole of $S^{n+1}$ and parallel transport this frame to all of the upper hemisphere $D_{+}$ (See Hatcher, "Vector bundles and K-theory", page 22 for a picture). Reflect along the equator to get another frame $E^- = (e^-_1, \cdots, e^-_{n+1})$ on the lower hemisphere $D_{-}$. One realizes $TS^{n+1}$ by identifying $D_{\pm} \times \Bbb R^{n+1}$ along the boundary $S^n \times \Bbb R^{n+1}$ by a fiberwise isomorphism $S^n \to \text{GL}(n+1)$ representing by change of basis from $E^+$ to $E^-$. This seems to be reflection (composed with other things) of $\Bbb R^{n+1}$ along varying hyperplanes, which is what $\mathcal{A}$ is. For $n = 1$, drawing the picture makes it clear that it's the degree $2$ map $S^1 \to SO(2)$, which is the clutching function of $TS^2$ as it has Euler class $\chi = 2$.
As conclusion, the antipodal embedding $S^n \to \Bbb R^{n+1}$ can be isotoped-through-immersions to the standard embedding if and only if $\alpha \in \pi_n SO(n+1)$ represented by the clutching function of $TS^{n+1}$ is trivial if and only if $S^{n+1}$ is parallelizable if and only if (by a theorem of Adams) $n = 0, 2, 6$. For the specific case of $n = 5$ since $\pi_4 SO(5) \cong \Bbb Z_2$, the standard embedding and the antipodal embedding constitute all the non-isotopic immersions of $S^4$ in $\Bbb R^5$.
EDIT: Here's a more direct way to see the result of the conclusion. Suppose $H : S^n \times I \to \Bbb R^{n+1}$ constitutes a regular homotopy between $H_0 = -\text{id}$ and $H_1 = \text{id}$. Consider the "movie" of this regular homotopy given by $G : S^n \times I \to \Bbb R^{n+1} \times I$, $G(x, t) = (H(x, t), t)$. Since the intro and the finale of this movie consists of embedded spheres, we can cap them off by the birth and death of the sphere (which contributes a $D^{n+1}$ each) to get a map $$\mathscr{I} : D^{n+1}_+ \cup S^n \times [0, 1] \cup D^{n+1}_{-} = S^{n+1} \to \Bbb R^{n+1} \times \Bbb R = \Bbb R^{n+2}$$ Since $\mathscr{I}$ is an immersion on each time slice $t = t_0$ and $\ker d\mathscr{I}$ does not contain the time direction, $\mathscr{I} : S^{n+1} \to \Bbb R^{n+2}$ is an immersion itself. If $\mathscr{G} : S^{n+1} \to S^{n+1}$ is the Gauss map corresponding to $\mathscr{I}$, and $\mathbf{t} \in S^{n+1}$ the unit vector in $\Bbb R^{n+2}$ pointing forward in the time dimension, $\mathscr{G}^{-1}(\mathbf{t})$ consists of exactly two points: the birth and the death of the movie. But since the movie of $\mathscr{I}$ depicts the birth of the identity embedding and death of the antipodal embedding, the local degree corresponding to these two points are opposite. Therefore $\text{deg} \,\mathscr{G} = -1 + 1 = 0$. Since the Gauss map has degree zero, $S^{n+1}$ is parallelizable.
I was interested to understand what $\pi_0 \text{Imm}(S^{n+1}, \Bbb R^{n+2})$ in this case might be (where $n$ is such that the sphere eversion of $S^n$ in $\Bbb R^{n+1}$ is possible). By Adam's theorem, since $S^{n+1}$ is parallelizable it has to be either $S^1$, $S^3$ or $S^7$ admit a natural multiplication as unit spheres in the normed division algebras $\Bbb C, \Bbb H$ and $\Bbb O$. Then the principal $SO(n+1)$-bundle $$SO(n+1) \to SO(n+2) \to S^{n+1}$$ admits a section by choosing the value of the section over the north pole of $S^{n+1}$ and then spreading it globally by multiplication to get a global section $S^{n+1} \to SO(n+2)$. This implies $SO(n+2) \simeq SO(n+1) \times S^{n+1}$ and therefore $\pi_0 \text{Imm}(S^{n+1}, \Bbb R^{n+2}) \cong \pi_{n+1} SO(n+2)$ is isomorphic to $\Bbb Z \oplus \pi_{n+1} SO(n+1)$.
Explicitly, there's a general map $I : \pi_0 \text{Imm}(S^m, \Bbb R^{m+1}) \to \pi_m S^m$ defined by $I([f]) = \pi \circ \mathcal{C}_f$ from composing the classifying map $\mathcal{C}_f : S^m \to SO(m+1)$ of $f$ with the fiber projection $\pi : SO(m+1) \to S^m$. If $\mathbf{e} = (e^1, \cdots, e^{m+1})$ is the identity frame of $\Bbb R^{m+1}$ and is considered as a constant framing on $S^m$, then for any immersion $f : S^m \to \Bbb R^{m+1}$ the corresponding classifying map $S^m\to SO(m+1) \simeq \text{Frame}_{m+1}(\Bbb R^{m+1})$ is $p \mapsto f_*(\mathbf{e}_p)$. $\pi : SO(m+1) \to S^m$ sends an $m+1$-frame to it's first vector component, so that we have the description $I([f])(p) = f_*(e^1_p)$. 
If $\Bbb R^{m+1}$ has a normed division algebra structure, then degree of the "framed Gauss map" $\mathscr{F} = \pi \circ \mathcal{C}_f$ of $f$ has a very natural relation with the degree of the normal Gauss map $\mathscr{G}$ of the immersion $f$. To work that out, notice that we have $\mathscr{F}(p) = f_*(e^1_p)$ and $\mathscr{G}(p) = f_*(\mathbf{n}_p)$. Notice that because of the product structure, $\mathbf{n}_p = p \bullet e^1_p$, which implies $\mathscr{G} = \text{id}\bullet\mathscr{F}$ where $\bullet$ denotes pointwise product of the self-maps of $S^m$. Using the lemma below, we can compute the degree  $$\boxed{\deg(\mathscr{G}) = \deg(\text{id}\bullet\mathscr{F}) = \deg(\mathscr{F}) + 1}$$
Lemma: $f, g : S^m \to S^m$, then $\text{deg}(f \bullet g) = \deg(f) + \deg(g)$ 
To see this, we use the cohomological definition of degree: Notice that 
$$f \bullet g : S^m \stackrel{\Delta}{\hookrightarrow} S^m \times S^m \stackrel{f\times g}{\to} S^m \times S^m \stackrel{\bullet}{\to} S^m$$ 
Let's call $1$ to be the generator of $H^m(S^m) \cong \Bbb Z$ and let $\alpha = H^m(f)(1)$ and $\beta = H^m(g)(1)$, and recall that $H^m(S^m \times S^m) \cong H^m(S^m) \oplus H^m(S^m) \cong \Bbb Z^2$ by Kunneth formula. $H^m(\Delta) : \Bbb Z^2 \to \Bbb Z$ is easily seen to be $(p, q) \mapsto p+q$, whereas $H^m(f \times g) : \Bbb Z^2 \to \Bbb Z^2$ is $(p, q) \mapsto (\alpha p, \beta q)$. Notice that $\bullet$ is identity restricted to $S^m \times \{e\}$ and $\{e\} \times S^m$ where $e$ is the identity for the normed division algebra $(\Bbb R^{m+1}, \bullet)$. This therefore implies $H^m(\bullet): \Bbb Z \to \Bbb Z^2$ is defined by $1 \mapsto (1, 1)$, the diagonal inclusion. In a nutshell, we get $H^m(f \bullet g)(1) = \alpha + \beta = H^m(f)(1) + H^m(g)(1)$ which concludes the proof, and establishes the boxed identity above.
This implies for $n=0,2,6$, the above map $\pi_0 \text{Imm}(S^{n+1}, \Bbb R^{n+2}) \to \pi_{n+1} S^{n+1}$ is simply given by $I([f]) = \deg(\mathscr{G}_f) - 1$. If $f$ is obtained from the movie procedure from sphere eversion of $S^n$ in $\Bbb R^{n+1}$, then $\mathscr{G}_f$ has degree zero, therefore $I([f]) = -1$. This implies that the movie immersion is not regularly homotopic to the standard immersion. For $n = 2$, the other factor in the isomorphism $\pi_3 SO(3) \cong \Bbb Z$ so that $\pi_0 \text{Imm}(S^3, \Bbb R^4) \cong \Bbb Z \oplus \Bbb Z$. In this paper of Hughes the other generator is apparently described as represented by the movie of the 2-sphere eversion done twice (birth-identity-antipodal-identity-death), which is kind of neat.
