Many theoretical accounts attribute much of the success of deep neural networks to feature learning, in which training changes the features represented by a network’s hidden layers. A theory of deep learning needs to explain which features a nonlinear network learns and how its weights represent them. We do not yet have a general theory that answers these questions.

We study this question as part of PIRAMID’s Advancements in Learning Theory team, which uses statistical physics to build foundations for mechanistic interpretability and AI safety. Our progress and plans describe a theory agenda centred on hidden weight distributions and covariances that can record learned structure. The mean field sequence introduces this viewpoint by treating each neuron as one member of a population whose distribution must be self consistent. The model below gives a solvable example in which we can derive the hidden weight distribution and its covariance.

Theorists have made progress by simplifying the task or the network. Rubin and collaborators give one of the clearest solvable examples in an equilibrium Bayesian model. They study a nonlinear network with two trainable layers in a teacher and student setting. They use statistical mechanics to derive the posterior over its weights. In that posterior, a hidden layer that starts isotropic acquires populations of weights aligned with one useful direction.

In this model, explicit feature learning is a change in the weight posterior. At the transition, the posterior over one scalar overlap develops new wells at finite overlap, and the hidden weight covariance gains a rank one term along the teacher direction. Rubin and collaborators connect the representation change to a first order transition in generalization and to grokking. The network can transfer from an easy linear target component to a harder cubic component.

We explain the feature learning result and work through its derivation. We begin with a cavity reduction, which isolates one neuron against the residual left by the others. The neuron’s DD dimensional posterior depends on one scalar overlap. We close the calculation with a self consistency equation that connects the learned weight distribution to the network’s prediction residual. We use the cavity calculation here because it exposes the population description in a simple model. We want to carry the representation and self consistency equations beyond the cavity construction.

Noa Rubin, Inbar Seroussi, and Zohar Ringel use the model to study grokking as a first order feature learning transition (Rubin, Seroussi & Ringel, 2024). A later paper by Rubin and collaborators extends the same mechanism across several network scaling regimes (Rubin et al., 2025). We follow the overlap transition from the first paper and use the adaptive kernel relation developed across both papers.

The target depends on one direction

Let the input be isotropic Gaussian,

x𝒩(0,ID), \mathbf x\sim\mathcal N(\mathbf 0,\mathbf I_D),

and let eD\mathbf e\in\RR^D be an unknown unit vector. The teacher depends only on the scalar projection

z=e𝖳x. z=\mathbf e^{\mathsf T}\mathbf x.

Its output combines the first and third probabilists’ Hermite polynomials,

y(x)=H1(z)+ϵH3(z)=z+ϵ(z33z).y(\mathbf x)=H_1(z)+\epsilon H_3(z) =z+\epsilon(z^3-3z).

The two modes are orthogonal under the standard Gaussian measure, and both use the same e\mathbf e. The coefficient ϵ\epsilon may be small.

The student is a network with two layers and an odd nonlinearity,

f(x)=i=1Naierf(wi𝖳x).f(\mathbf x)=\sum_{i=1}^{N}a_i\,\operatorname{erf}(\mathbf w_i^{\mathsf T}\mathbf x).

Each hidden weight wi\mathbf w_i can rotate. We measure its alignment with the teacher by the overlap

ρi=wi𝖳e. \rho_i=\mathbf w_i^{\mathsf T}\mathbf e.

Plate 1.1 shows the task and the scalar order parameter.

Plate 1.1 One direction carries both target modes

one informative projection

Gaussian inputs projected onto one teacher directionA cloud of input points surrounds the origin. A single diagonal axis marked e crosses the cloud, and faint perpendicular segments show several projections onto it.e\mathbf ez=e𝖳xz=\mathbf e^{\mathsf T}\mathbf x

two orthogonal target modes

Linear and cubic Hermite modes of the teacherA straight green line represents the first Hermite mode. A wine cubic curve represents the third Hermite mode. Both depend on the same scalar coordinate z.H1(z)=zH_1(z)=zH3(z)=z33zH_3(z)=z^3-3zzz

a neuron can rotate toward it

Hidden weights described by their overlap with the teacherQuiet radial directions surround an origin. A highlighted hidden weight approaches the teacher axis. Their scalar alignment is rho.e\mathbf ew\mathbf wρ=w𝖳e\rho=\mathbf w^{\mathsf T}\mathbf e
The teacher uses one scalar projection. The linear and cubic Hermite modes share that coordinate. Feature learning appears when the hidden weight distribution acquires finite overlap with the teacher direction.

The linear part contains enough information to estimate the teacher direction. Consider

e^=1Pμ=1Pyμxμ.\widehat{\mathbf e} =\frac1P\sum_{\mu=1}^{P}y^\mu\mathbf x^\mu.

Gaussian orthogonality gives

𝔼[yx]=e𝔼[zH1(z)]+ϵe𝔼[zH3(z)]=e. \EE[y\mathbf x] =\mathbf e\,\EE[zH_1(z)] +\epsilon\mathbf e\,\EE[zH_3(z)] =\mathbf e.

The squared estimation error is O(D/P)O(D/P), so P=O(D)P=O(D) examples identify e\mathbf e to order one accuracy. The linear term supplies the direction. Without it, this derivation does not give the network a route to the cubic target. Another mechanism could still learn the target.

After O(D)O(D) samples, the data contain the direction. We need to explain when training puts that direction into the hidden weight distribution and how the new distribution changes prediction.

A Bayesian model exposes the hidden representation

The tractable model treats training as Bayesian inference. Put independent priors on the hidden and readout weights,

wi𝒩 ⁣(0,σw2DID),ai𝒩(0,sa2), \mathbf w_i\sim\mathcal N\!\left(\mathbf 0,\frac{\sigma_w^2}{D}\mathbf I_D\right), \qquad a_i\sim\mathcal N(0,s_a^2),

and use a Gaussian likelihood with variance TT on a training set of size PP,

p(𝒟{ai,wi})exp ⁣[12Tμ=1P(yμf(xμ))2]. p(\mathcal D\mid\{a_i,\mathbf w_i\}) \propto \exp\!\left[-\frac1{2T}\sum_{\mu=1}^{P} \bigl(y^\mu-f(\mathbf x^\mu)\bigr)^2\right].

Standard scaling takes sa2=O(1/N)s_a^2=O(1/N). The posterior balances a prior cost in DD dimensions against the fit reward from PP examples.

The hidden weight distribution provides a direct measure of feature learning. Feature learning appears when that distribution changes from isotropic to aligned with a direction selected by the target. Equilibrated Langevin training samples this posterior. Gradient descent can follow another path and can take a different time to reach equilibrium, so its dynamics need a separate argument.

Score one neuron against the residual

Remove one neuron from the network and call the residual left by the other N1N-1 neurons

rμ=yμf(xμ). r_\mu=y^\mu-f_{-}(\mathbf x^\mu).

For a candidate hidden weight w\mathbf w, define its feature vector on the training set by

ϕμ(w)=erf(w𝖳xμ). \phi_\mu(\mathbf w)=\operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x^\mu).

The selected neuron contributes aϕa\boldsymbol\phi. Conditional on w\mathbf w, its readout coefficient is Gaussian, so we can integrate it exactly. Introduce

S(w)=ϕ𝖳r,Q(w)=ϕ𝖳ϕ,L(w)=1+sa2TQ(w). S(\mathbf w)=\boldsymbol\phi^{\mathsf T}\mathbf r, \qquad Q(\mathbf w)=\boldsymbol\phi^{\mathsf T}\boldsymbol\phi, \qquad L(\mathbf w)=1+\frac{s_a^2}{T}Q(\mathbf w).

Completing the square in aa gives

p(wr,X)p0(w)L(w)1/2exp ⁣[sa2S(w)22T2L(w)].p(\mathbf w\mid\mathbf r,X) \propto p_0(\mathbf w)\,L(\mathbf w)^{-1/2} \exp\!\left[ \frac{s_a^2S(\mathbf w)^2}{2T^2L(\mathbf w)} \right].

Equation is exact for a fixed cavity residual. The prior keeps w\mathbf w small and isotropic. The factor L1/2L^{-1/2} accounts for the cost of fitting a readout coefficient, and the exponential rewards a feature that correlates with the residual. The cavity step holds the residual fixed while it scores one neuron. The final self consistency step returns the neuron to the network and requires the residual and adaptive kernel to agree.

Calculation Complete the square in the readout weight

Expand the squared residual,

raϕ2=r22aS(w)+a2Q(w). \norm{\mathbf r-a\boldsymbol\phi}^2 =\norm{\mathbf r}^2-2aS(\mathbf w)+a^2Q(\mathbf w).

The terms that depend on aa have the form

A(w)2a2+B(w)a,A(w)=1sa2+Q(w)T,B(w)=S(w)T. -\frac{A(\mathbf w)}2a^2+B(\mathbf w)a, \qquad A(\mathbf w)=\frac1{s_a^2}+\frac{Q(\mathbf w)}T, \qquad B(\mathbf w)=\frac{S(\mathbf w)}T.

The Gaussian identity

exp ⁣(A2a2+Ba) ⁣da=2πAexp ⁣(B22A) \int_{-\infty}^{\infty} \exp\!\left(-\frac A2a^2+Ba\right)\dd a =\sqrt{\frac{2\pi}{A}}\exp\!\left(\frac{B^2}{2A}\right)

then gives Equation after removing factors that do not depend on w\mathbf w.

The fit reward depends on S2S^2, so the signs w\mathbf w and w-\mathbf w remain equivalent. The mean weight can stay at zero while an even moment such as 𝔼[ρ2]\EE[\rho^2] records direction learning.

Reduce the hidden weight posterior to one overlap

Before the transition, the residual lies mainly in the same two Hermite modes as the teacher. Write its population form as

r(x)=bH1(z)+cH3(z).r(\mathbf x)=bH_1(z)+cH_3(z).

For one hidden weight, let

g=w𝖳x,q=w2,ρ=w𝖳e. g=\mathbf w^{\mathsf T}\mathbf x, \qquad q=\norm{\mathbf w}^2, \qquad \rho=\mathbf w^{\mathsf T}\mathbf e.

The pair (g,z)(g,z) is jointly Gaussian, with covariance ρ\rho. For a smooth function FF, the higher order Stein identity gives

𝔼[F(g)Hk(z)]=ρk𝔼[F(k)(g)].\EE[F(g)H_k(z)] =\rho^k\EE[F^{(k)}(g)].

The identity turns the kkth teacher mode into the kkth power of the overlap. Gaussian inputs give the clean ρk\rho^k factor. With non Gaussian inputs, higher moments can mix the teacher modes, so the calculation needs a different moment expansion.

Calculation Derive the Gaussian Hermite identity

Write the correlated Gaussian variables as

g=ρz+qρ2ξ,ξ𝒩(0,1),ξ independent of z. g=\rho z+\sqrt{q-\rho^2}\,\xi, \qquad \xi\sim\mathcal N(0,1), \qquad \xi\ \text{independent of }z.

For fixed ξ\xi, set h(z)=F(g)h(z)=F(g). The one variable Stein identity gives

𝔼[Hk(z)h(z)]=𝔼[h(k)(z)]. \EE[H_k(z)h(z)]=\EE[h^{(k)}(z)].

Each derivative with respect to zz contributes one factor of ρ\rho, so

h(k)(z)=ρkF(k)(g). h^{(k)}(z)=\rho^kF^{(k)}(g).

Averaging over ξ\xi gives Equation . For F(g)=erf(g)F(g)=\operatorname{erf}(g), the first and third derivative averages are

𝔼[F(g)]=2π11+2q,𝔼[F(g)]=4π1(1+2q)3/2. \EE[F'(g)] =\frac{2}{\sqrt\pi}\frac1{\sqrt{1+2q}}, \qquad \EE[F'''(g)] =-\frac{4}{\sqrt\pi}\frac1{(1+2q)^{3/2}}.

Setting F=erfF=\operatorname{erf} yields the two correlations needed here,

𝔼[erf(g)H1(z)]=2ρπ(1+2q),𝔼[erf(g)H3(z)]=4ρ3π(1+2q)3/2. \begin{aligned} \EE[\operatorname{erf}(g)H_1(z)] &=\frac{2\rho}{\sqrt{\pi(1+2q)}},\\ \EE[\operatorname{erf}(g)H_3(z)] &=-\frac{4\rho^3}{\sqrt\pi(1+2q)^{3/2}}. \end{aligned}

The resulting correlation between the feature and residual is

m(ρ,q)=𝔼[r(x)erf(g)]=2πρ1+2q(b2cρ21+2q).m(\rho,q) =\EE[r(\mathbf x)\operatorname{erf}(g)] =\frac{2}{\sqrt\pi} \frac{\rho}{\sqrt{1+2q}} \left( b-\frac{2c\rho^2}{1+2q} \right).

The feature variance is

v(q)=𝔼[erf(g)2]=2πarcsin ⁣(2q1+2q).v(q)=\EE[\operatorname{erf}(g)^2] =\frac{2}{\pi}\arcsin\!\left(\frac{2q}{1+2q}\right).

For large PP, the empirical quantities concentrate as SPmS\simeq Pm and QPvQ\simeq Pv.

Calculation Check the sample sum approximation

For fixed w\mathbf w and residual function rr, both S/PS/P and Q/PQ/P average PP independent terms. Their variances are

Var ⁣[S(w)P]=1PVar ⁣[r(x)erf(w𝖳x)], \Var\!\left[\frac{S(\mathbf w)}P\right] =\frac1P\Var\!\left[r(\mathbf x)\operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x)\right],

and

Var ⁣[Q(w)P]=1PVar ⁣[erf(w𝖳x)2]. \Var\!\left[\frac{Q(\mathbf w)}P\right] =\frac1P\Var\!\left[\operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x)^2\right].

Both averages fluctuate on the scale P1/2P^{-1/2}. Under sa2N1s_a^2\sim N^{-1} and PNDP\sim N\sim D, the leading log posterior is O(D)O(D) while the first correction for fixed w\mathbf w is O(D)O(\sqrt D).

The components of w\mathbf w orthogonal to e\mathbf e also concentrate, so

qσw2+ρ2. q\simeq\sigma_w^2+\rho^2.
Calculation Check the orthogonal norm concentration

Decompose the weight as

w=ρe+w,e𝖳w=0. \mathbf w=\rho\mathbf e+\mathbf w_\perp, \qquad \mathbf e^{\mathsf T}\mathbf w_\perp=0.

Under the Gaussian prior, the D1D-1 orthogonal coordinates have variance σw2/D\sigma_w^2/D. Therefore

w2σw2DχD12. \norm{\mathbf w_\perp}^2 \sim\frac{\sigma_w^2}{D}\chi^2_{D-1}.

Its mean approaches σw2\sigma_w^2, and its relative fluctuations have size D1/2D^{-1/2}.

After these approximations, Equation gives the scalar distribution

p(ρb,c)eU(ρ;b,c),p(\rho\mid b,c)\propto e^{-U(\rho;b,c)},

with

U(ρ;b,c)=Dρ22σw2+12logL(ρ)sa2P2m(ρ)22T2L(ρ),L(ρ)=1+sa2PTv(σw2+ρ2).U(\rho;b,c) =\frac{D\rho^2}{2\sigma_w^2} +\frac12\log L(\rho) -\frac{s_a^2P^2m(\rho)^2}{2T^2L(\rho)}, \qquad L(\rho)=1+\frac{s_a^2P}{T}v(\sigma_w^2+\rho^2).

The first term is the prior cost of selecting one direction in DD dimensions. The last term rewards a feature that fits the residual. Since m(ρ)m(\rho) begins at first order in ρ\rho, enough data can make weights with finite overlap competitive.

Proposition (Alignment scale)

Let sa2=O(1/N)s_a^2=O(1/N) and suppose bb, TT, and the prior scales remain order one. Near ρ=0\rho=0, the fit reward competes with the directional prior cost when

P2N=O(D). \frac{P^2}{N}=O(D).

Hence Palign=O(ND)P_{\mathrm{align}}=O(\sqrt{ND}). In the proportional regime N=O(D)N=O(D), the posterior can acquire finite overlap at P=O(D)P=O(D).

Calculation Expand the action near zero overlap

Write

m(ρ)=αρ+O(ρ3),L(ρ)=L0+ρ2+O(ρ4). m(\rho)=\alpha\rho+O(\rho^3), \qquad L(\rho)=L_0+\ell\rho^2+O(\rho^4).

Then

m(ρ)2=α2ρ2+O(ρ4),logL(ρ)=logL0+L0ρ2+O(ρ4). m(\rho)^2=\alpha^2\rho^2+O(\rho^4), \qquad \log L(\rho)=\log L_0+\frac{\ell}{L_0}\rho^2+O(\rho^4).

Substitution into the scalar potential gives

U(0)=Dσw2+L0sa2P2α2T2L0. U''(0) =\frac{D}{\sigma_w^2} +\frac{\ell}{L_0} -\frac{s_a^2P^2\alpha^2}{T^2L_0}.

The centered state loses local stability when this curvature changes sign. With sa2=O(N1)s_a^2=O(N^{-1}), the prior and fit terms balance at P=O(ND)P=O(\sqrt{ND}).

The curvature test locates a spinodal boundary. The first order transition studied by Rubin, Seroussi, and Ringel is a global statement. Wells at finite overlap can reach the same depth as the centered well before the center loses local stability (Rubin, Seroussi & Ringel, 2024). New minima at finite overlap first appear, then reach the same depth as the centered minimum, and later make the center unstable. The middle event is the first order transition.

The 2024 paper combines the sample count and the remaining scales into an effective interaction,

u=P2sa2T2DN,U~u(ρ)D=ρ22σw22uπρ21+2(σw2+ρ2)(b2cρ21+2(σw2+ρ2))2.u=\frac{P^2s_a^2}{T^2DN}, \qquad \frac{\widetilde U_u(\rho)}{D} =\frac{\rho^2}{2\sigma_w^2} -\frac{2u}{\pi} \frac{\rho^2}{1+2(\sigma_w^2+\rho^2)} \left( b-\frac{2c\rho^2}{1+2(\sigma_w^2+\rho^2)} \right)^2.

Move the control in Plate 4.1 to follow the conditional action as uu increases. The adjacent hidden population uses the same overlap coordinate and shows how posterior mass moves between the wells.

Plate 4.1 Aligned neuron populations appear as new posterior wells
Effective load uu29.45 Equal-depth coexistence. Population phase GMFL-I. Population second moment R equals 0.15.
The overlap action as effective load changesA line plot of the normalized overlap action. Move the effective-load control to see symmetric side wells appear, dominate, and destabilize the centred state. equal-depth coexistenceUu(ρ)minUuU_u(\rho)-\min U_u1-10011ρ\rho

Drag, or use the arrow keys for fine steps.

Move the effective interaction uu through an analytic slice of the 2024 mean field action. The curve fixes σw2=0.5\sigma_w^2=0.5, b=0.30b=0.30, and c=0.30c=-0.30. The population of forty units follows the same minima. Units stay near zero in GFL, divide between the centre and side wells in GMFL-I, and occupy the side wells in GMFL-II. Vertical position separates the dots. The displayed RR is the second moment of the dots. The dots and their mixture weights are illustrative. The full theory solves bb and cc through self consistency.

The hidden covariance records the learned direction

Let

R=𝔼[ρ2]. R=\EE[\rho^2].

Before alignment, R=O(D1)R=O(D^{-1}). Once the posterior assigns finite mass near ρ=±ρ\rho=\pm\rho_*, RR becomes O(1)O(1). Symmetry still gives 𝔼[ρ]=0\EE[\rho]=0.

Decompose a hidden weight into parts parallel and orthogonal to the teacher,

w=ρe+ξ,e𝖳ξ=0. \mathbf w=\rho\mathbf e+\boldsymbol\xi, \qquad \mathbf e^{\mathsf T}\boldsymbol\xi=0.

The orthogonal directions remain equivalent. To leading order at large DD,

𝔼[ww𝖳]=σw2D1(Iee𝖳)+Ree𝖳.\EE[\mathbf w\mathbf w^{\mathsf T}] =\frac{\sigma_w^2}{D-1} \bigl(\mathbf I-\mathbf e\mathbf e^{\mathsf T}\bigr) +R\,\mathbf e\mathbf e^{\mathsf T}.

The covariance receives a rank one correction along e\mathbf e. Plate 5.1 shows why the second moment is the right measure. A symmetric pair of aligned populations can have zero mean while its covariance records the learned direction.

Plate 5.1 The covariance records direction learning

before direction learning

An isotropic hidden-weight distributionHidden weights form an approximately circular cloud around zero in a plane containing the teacher direction.e\mathbf eR=𝔼[ρ2]=O(D1)R=\EE[\rho^2]=O(D^{-1})

after direction learning

A hidden-weight distribution elongated along the teacherThe cloud is symmetric about zero but elongated along the teacher axis, showing a rank-one covariance increase without a nonzero mean weight.e\mathbf eR=𝔼[ρ2]=O(1)R=\EE[\rho^2]=O(1)
𝔼[ww𝖳]=σw2D1I+(Rσw2D1)ee𝖳\EE[\mathbf w\mathbf w^{\mathsf T}]=\frac{\sigma_w^2}{D-1}\mathbf I+\left(R-\frac{\sigma_w^2}{D-1}\right)\mathbf e\mathbf e^{\mathsf T}
The weight mean can remain zero because the two signs are equivalent. The second moment acquires an eigenvalue of order one along the teacher direction. Multiplying the original isotropic kernel by one scalar cannot represent this directional change.

The rank one term is a direct measure of explicit feature learning. A scalar rescaling changes every direction by the same factor, while the adaptive covariance changes the target direction. Rubin and collaborators retain this directional information across several network scaling regimes, including regimes where a scalar rescaling predicts some mean outputs well (Rubin et al., 2025).

Self consistency closes the feature learning loop

The one neuron posterior starts from coefficients (b,c)(b,c) in the residual. The whole network determines those coefficients through its current prediction.

For a candidate residual, the overlap posterior defines an adaptive kernel,

Kμν(b,c)=Nsa2𝔼p(wb,c) ⁣[erf(w𝖳xμ)erf(w𝖳xν)].K_{\mu\nu}(b,c) =Ns_a^2\, \EE_{p(\mathbf w\mid b,c)}\!\left[ \operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x^\mu) \operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x^\nu) \right].

For that kernel, the posterior mean residual on the training set is

r=T(K(b,c)+TI)1y.\overline{\mathbf r} =T\bigl(\mathbf K(b,c)+T\mathbf I\bigr)^{-1}\mathbf y.

Projecting r\overline{\mathbf r} onto H1H_1 and H3H_3 gives new coefficients (b,c)(b,c). A solution is a fixed point of this map. Plate 6.1 shows the full loop.

Plate 6.1 The learned features and residual must agree
The adaptive-kernel self-consistency loopFour states form a clockwise loop. Residual coefficients determine an overlap posterior, the posterior determines an adaptive kernel, the kernel determines a predictor, and the predictor returns new residual coefficients.r(x)=bH1(z)+cH3(z)r(x)=bH_1(z)+cH_3(z)p(ρb,c)p(\rho\mid b,c)Kb,cK_{b,c}rˉ=T(Kb,c+TI)1y\bar r=T(K_{b,c}+T\mathbf I)^{-1}yfit one neuron\text{fit one neuron}average features\text{average features}kernel prediction\text{kernel prediction}project residual\text{project residual}
The residual sets the posterior over hidden directions. The posterior sets the adaptive kernel, and the kernel changes the residual. A self consistent solution satisfies all parts of the loop together.

The calculation for one neuron becomes a network theory only after solving this loop. The residual specifies what remains to be learned. The hidden weight distribution changes the kernel, and the new kernel produces a new residual. We want to carry the requirement that representation and prediction error agree beyond this model.

A sample complexity corollary

The learned direction changes the kernel power assigned to the cubic teacher mode. Before alignment, a random hidden direction has ρ=O(D1/2)\rho=O(D^{-1/2}), so the cubic kernel power is O(D3)O(D^{-3}). After alignment, a finite part of the posterior has ρ=O(1)\rho=O(1), so the same channel has order one strength.

Calculation Derive the cubic sample scale

The third Hermite coefficient of one erf feature is

h3(w)=𝔼[erf(w𝖳x)H3(z)]=4ρ3π(1+2q)3/2. h_3(\mathbf w) =\EE[\operatorname{erf}(\mathbf w^{\mathsf T}\mathbf x)H_3(z)] =-\frac{4\rho^3}{\sqrt\pi(1+2q)^{3/2}}.

At initialization, ρ=O(D1/2)\rho=O(D^{-1/2}), which gives

h3(w)=O(D3/2),h3(w)2=O(D3). h_3(\mathbf w)=O(D^{-3/2}), \qquad h_3(\mathbf w)^2=O(D^{-3}).

The squared coefficient controls the kernel power. With Nsa2=O(1)Ns_a^2=O(1), averaging over width leaves the D3D^{-3} scale unchanged. A fixed kernel needs P=O(D3)P=O(D^3) examples before the cubic mode has order one signal relative to noise. Once the posterior has finite overlap, h3(w)=O(1)h_3(\mathbf w)=O(1).

The linear target mode reveals e\mathbf e at P=O(D)P=O(D). The adaptive network can reuse the learned direction for the cubic function of the same coordinate, while a fixed kernel still needs P=O(D3)P=O(D^3). The exponent three comes from squaring h3ρ3h_3\propto\rho^3 when a random overlap has size ρ=O(D1/2)\rho=O(D^{-1/2}). It does not come from counting every cubic monomial in DD variables.

Plate 7.1 compares the two sample scales.

Plate 7.1 A learned direction changes the cubic sample scale
Two routes to the cubic teacher modeThe upper route keeps random hidden directions. Their overlap is of order one over square root D, so cubic kernel power is of order D to the minus three and sample complexity is D cubed. The lower route uses the linear teacher component to learn the direction with D samples, after which cubic power is order one.fixed featuresadaptive featureswp0\mathbf w\sim p_0ρ=O(D1/2)\rho=O(D^{-1/2})h32=O(D3)h_3^2=O(D^{-3})Pfixed=O(D3)P_{\mathrm{fixed}}=O(D^3)H1 estimates eH_1\ \text{estimates }\mathbf ePalign=O(D)P_{\mathrm{align}}=O(D)ρ=O(1),h32=O(1)\rho=O(1),\quad h_3^2=O(1)Padaptive=O(D)P_{\mathrm{adaptive}}=O(D)

fixed features

  1. wp0\mathbf w\sim p_0
  2. ρ=O(D1/2)\rho=O(D^{-1/2})
  3. h32=O(D3)h_3^2=O(D^{-3})
  4. Pfixed=O(D3)P_{\mathrm{fixed}}=O(D^3)

adaptive features

  1. H1 estimates eH_1\ \text{estimates }\mathbf e
  2. Palign=O(D)P_{\mathrm{align}}=O(D)
  3. ρ=O(1),h32=O(1)\rho=O(1),\quad h_3^2=O(1)
  4. Padaptive=O(D)P_{\mathrm{adaptive}}=O(D)
For fixed features, cubic kernel power falls as the inverse cube of dimension, so the required sample count grows as the cube of dimension. The adaptive network uses the linear mode to identify the teacher direction with a sample count proportional to dimension. Finite overlap then gives the cubic mode order one strength along the same direction.

The later adaptive theory predicts learning of the nonlinear component at P=O(D)P=O(D), while the NNGP and scalar rescaling descriptions retain the O(D3)O(D^3) requirement (Rubin et al., 2025). The earlier grokking paper observes the same change after the feature learning transition (Rubin, Seroussi & Ringel, 2024).

What we are working on

The calculation above tells us when a hidden layer learns a useful direction, but it leaves open how the layer distributes what it has learned. If ρi\rho_i is neuron ii’s overlap with the teacher direction, then its contribution to a degree-kk feature is proportional to aiρika_i\rho_i^k. The same total feature can come from a few strongly aligned neurons or from many neurons whose individual contributions are small. A small aligned population stores the feature in specialists. A broad population stores the feature across many units.

Both populations can compute the same function while having different hidden covariance and unit selectivity. They can also respond differently when neurons are removed. A theory that predicts only the output cannot distinguish them. We want a theory that begins with the posterior over weights and predicts how many neurons participate and how strongly each contributes.

Our current work asks what controls the division between distributed and specialized representations. We study simple models in which training can store the same function in different ways, and we use the posterior over weights to track how many neurons participate. A theory of representation is necessary for interpretability because a faithful account must describe the function a model computes and how its neurons represent that function.

References

  1. Noa Rubin, Inbar Seroussi, and Zohar Ringel (). Grokking as a First Order Phase Transition in Two Layer Networks. International Conference on Learning Representations. OpenReview:3ROGsTX3IR. ↩1 ↩2 ↩3
  2. Noa Rubin, Kirsten Fischer, Javed Lindner, Inbar Seroussi, Zohar Ringel, Michael Krämer, and Moritz Helias (). From Kernels to Features: A Multi-Scale Adaptive Theory of Feature Learning. Proceedings of the 42nd International Conference on Machine Learning, PMLR 267, 52225-52257. PMLR:v267/rubin25a. ↩1 ↩2 ↩3