SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\operatorname {e r r} (h) = \int_ {X} p (x) | | f (x) - h (x) | | _ {2} ^ {2} d x = \int_ {x \in F} \int_ {z \in G x} p (z) | | f (z) - h (z) | | _ {2} ^ {2} d z d x.
$$
Denote  $e(x) = \int_{Gx} p(z) \| f(z) - h(z) \|_2^2 dz$ . Since  $h$  is  $G$ -invariant, there exists  $c \in \mathbb{R}^n$  such that  $h(z) = c$  for all  $z \in Gx$ . Then  $e(x)$  can be written as  $e(x) = \int_{Gx} p(z) \| f(z) - c \|_2^2 dz$ . Taking the derivative of  $e(x)$  with respect to  $c$  and setting it to 0 gives  $c^*$ , the minimum of  $e(x)$ ,  $c^* = \frac{\int_{Gx} p(z) f(z) dz}{\int_{Gx} p(z) dz} = \mathbb{E}_{Gx}[f]$ . Substituting  $c^*$  into  $e(x)$  we have
$$
e (x) \geq \int_ {G x} p (G x) \frac {p (z)}{p (G x)} | | \mathbb {E} _ {G x} [ f ] - f (z) | | _ {2} ^ {2} d z = p (G x) \mathbb {V} _ {G x} [ f ].
$$
We can obtain the claim by taking the integral of  $e(x)$  over the fundamental domain  $F$ .
4.3 Lower Bound for Equivariant Regression
We now prove a lower bound for  $\operatorname{err}(h)$  in a regression task given the model  $h$  is equivariant, that is,  $h(\rho_X(g)x) = \rho_Y(g)h(x)$  where  $g \in G$ ,  $\rho_X$  and  $\rho_Y$  are group representations associated with  $X$  and  $Y$ . We will denote  $\rho_X(g)x$  and  $\rho_Y(g)y$  by  $gx$  and  $gy$ , leaving the representation implicit. Assume  $Y = \mathbb{R}^n$  and  $\alpha(x,g)$  is the same as in equation 3. Let  $\mathrm{Id}$  be the identity. Define a matrix  $Q_{Gx} \in \mathbb{R}^{n \times n}$  and  $q(gx) \in \mathbb{R}^{n \times n}$  so that  $\int_G q(gx) dg = \mathrm{Id}$  by
$$
Q _ {G x} = \int_ {G} p (g x) \rho_ {Y} (g) ^ {T} \rho_ {Y} (g) \alpha (x, g) d g, \quad q (g x) = Q _ {G x} ^ {- 1} p (g x) \rho_ {Y} (g) ^ {T} \rho_ {Y} (g) \alpha (x, g). \tag {5}
$$
(b)
Figure 5: An example regression task. (a) The value of  $f(x)$  and the transformation rule (purple) with respect to group  $G = C_4$  for all  $x \in X$ . The four points belong to a single orbit. (b) When using an invariant network, the minimal error (red) is obtained when the invariant network outputs the mean value (green) of the orbit. (c) For an equivariant network, the minimizer (green) can be obtained by taking the mean of the  $G$ -stabilized  $f(x)$  (inversely transformed) (blue) for all  $x$  in the orbit with respect to the transformation rule in the orbit. (d) The minimal error of an equivariant network.
Here, for simplicity, we assume  $Q_{Gx}$  is an invertible matrix. (See Appendix D for general case).
If  $f$  is equivariant,  $g^{-1}f(gx)$  is a constant for all  $g\in G$ . Define  $\mathbf{E}_G[f,x]$
$$
\mathbf {E} _ {G} [ f, x ] = \int_ {G} q (g x) g ^ {- 1} f (g x) d g. \tag {6}
$$
Theorem 4.9. The error of  $h$  has lower bound  $\operatorname{err}(h) \geq \int_{F} \int_{G} p(gx) \| f(gx) - g\mathbf{E}_{G}[f, x] \|_{2}^{2} \alpha(x, g) dgdx$ .
See Appendix D for the proof. Intuitively,  $\mathbf{E}_G[f,x]$  is the minimizer obtained by taking the mean of all inversely transformed  $f(x)$  for all  $x$  in the orbit, see Figure 5cd and Example 4.11 below.
Corollary 4.10. Denote  $p(Gx) = \int_{Gx} p(z) dz$ . Denote  $q_x : g \mapsto q(gx)$ . Define  $G$ -stabilized  $f$  as  $f_x : g \mapsto g^{-1} f(gx)$ . When  $\rho_Y$  is an orthogonal representation  $\rho_Y : G \to \mathrm{O}(n) \subset GL(n)$ ,  $q_x$  is a probability density function on  $G$ . Denote the variance of  $f_x$  as  $\mathbb{V}_G[f_x]$  where  $g \sim q_x$ . The error has a lower bound  $\operatorname{err}(h) \geq \int_F p(Gx) \mathbb{V}_G[f_x] dx$ .
See Appendix E for the proof. Notice that Corollary 4.10 is a generalization of Theorem 4.8. That is, Theorem 4.8 can be recovered by taking  $\rho_Y(g) = \mathrm{Id}$  (See the proof in Appendix F).
Example 4.11 (Lower bound example of a regression task). Consider a regression problem where  $X = \{x_0, x_1, x_2, x_3\}$  and  $Y = \mathbb{R}^2$ . Assume  $p$  is uniform density. The cyclic group  $G = C_4 = \{e, r, r^2, r^3\}$  (where  $e = 0$  rotation and  $r = \pi/2$  rotation) acts on  $X$  through  $x_1 = rx_0$ ;  $x_2 = rx_1$ ;  $x_3 = rx_2$ ;  $x_0 = rx_3$  (i.e., there is only one orbit  $Gx = X$ ).  $g \in G$  acts on  $y \in Y$  through  $\rho_Y(g) = \left( \frac{\cos g - \sin g}{\sin g \cos g} \right)$ . Figure 5a shows the output of  $f(x), \forall x \in X$ . First, consider a  $G$ -invariant network  $h$ . Since there is only one orbit, Theorem 4.8 can be simplified as:  $\operatorname{err}(h) \geq \mathbb{V}_X[f]$ , the variance of  $f$  over  $X$ . This can be calculated by first taking the mean of  $f(x)$  then calculating the mean square error (MSE) from all  $x$  to the mean (Figure 5b). Second, consider a  $G$ -equivariant network  $h$ . Since  $G$  is discrete,  $gx$  permutes the order of  $X$ ,  $\rho_Y$  is an orthogonal representation, and there is only one orbit, Corollary 4.10 can be written as  $\operatorname{err}(h) \geq \mathbb{V}_G[f_x]$ , the variance of  $G$ -stabilized  $f$ . First, to calculate  $\mathbb{E}_G[f_x]$ , let  $x = x_0$ , we stabilize  $g$  from  $f$  by  $g^{-1}f(gx)$  for all  $g \in G$ , then take the mean (Figure 5c). We can then find  $\mathbb{V}_G[f_x]$  by calculating the MSE between  $f(x)$  and transformed mean  $g\mathbb{E}_G[f_x]$  (Figure 5d). Appendix H.3 shows an experiment in this example's environment.
5 Harmful Extrinsic Equivariance
Wang et al. [58] demonstrate that extrinsic equivariance, where the symmetry imposed on the model leads to out-of-distribution data with respect to the input distribution, can lead to a higher performance on the original training data. In this section, we argue that this is not necessarily true in all cases, and there can exist scenarios where extrinsic equivariance can even be harmful to the learning problem.
Consider a binary classification task where the domain is discrete and contains only a set of four points  $S \subset \mathbb{R}^3$ , and their labels are either  $\{-1, +1\}$  as shown in Figure 6a. We consider the probability density  $p$  to be uniform for this domain, i.e.,  $p(x) = 1/4$  for the four points  $S$ , and  $p = 0$  elsewhere.
This domain is used for both model training and testing so there is no distribution shift. We consider two model classes,  $\mathcal{F}_N$ , the set of all linear models, and  $\mathcal{F}_E$ , the set of all linear models which are invariant with respect to the cyclic group  $C_2 = \{1, g\}$ , where  $g(x_1, x_2, x_3) = (x_1, x_2, -x_3)$ .  $\mathcal{F}_N$  corresponds to an unconstrained or a non-equivariant model class and  $\mathcal{F}_E$  corresponds to an extrinsically equivariant class for this domain. For the labeling shown in Figure 6, the hyperplane  $x_3 = 0$  correctly classifies all samples and is contained in  $\mathcal{F}_N$ . However, a function  $f_e \in \mathcal{F}_E$  is equivalent to a linear classifier on  $\mathbb{R}^2$  and effectively sees the data as Figure 6b<sup>4</sup>. This exclusive-or problem does not admit a linear solution (it can be correct for at most 3 points).
Concretely, we can compute the empirical Rademacher complexity, a standard measure of model class expressivity, for nonequivariant and extrinsically equivariant model classes and show that  $\mathcal{F}_E$  has lower complexity than  $\mathcal{F}_N$ . Recall that empirical Rademacher complexity is defined as  $\Re_S(\mathcal{F}) = \mathbb{E}_{\sigma}\left[\sup_{f\in \mathcal{F}}\frac{1}{m}\sum_{i = 1}^{m}\sigma_if(x^i)\right]$ , where  $S$  is the set of  $m$  samples and  $\sigma = (\sigma_1,\dots ,\sigma_m)^\top$ ,  $\sigma_i\in \{-1, + 1\}$  are independent uniform Rademacher random variables, and  $x^{i}$  is the  $i$ -th sample. As there exists some linear function  $f_{n}\in \mathcal{F}_{N}$  that fully classifies  $S$  for any combination of labels,  $\Re_S(\mathcal{F}_N) = 1$ . For the extrinsic equivariance case, of the 16 possible label combinations, there are two cases where  $f_{e}\in \mathcal{F}_{E}$  can at most classify 3 out of 4 points correctly, and thus  $\Re_S(\mathcal{F}_E) = \frac{31}{32} < \Re_S(\mathcal{F}_N)$  (see Appendix G
(a) Data
(b) Transformed (c)  $C_2$ -equivariant data view
Figure 6: An example dataset where extrinsic equivariance increases the problem difficulty. The samples are of the form  $x = (x_{1},x_{2},x_{3})$  and the labels are shown as different shapes. A  $C_2$ -equivariant linear model transforms the original data (a) into (b), which is equivalent to viewing the data as in (c). The original task has an easy solution (e.g. hyperplane at  $x_{3} = 0$ ), while the  $C_2$ -invariant view is the classic exclusive-or problem.
for the calculations). This illustrates that in certain cases, extrinsic equivariance can lead to lower model expressivity than no equivariance and thus be harmful to learning.
We perform experiments to validate our theoretical analysis on both the lower bounds (Section 4) and the harmful extrinsic equivariance (Section 5). We find that our bounds accurately predict empirical model error. In addition to the experiments in this section, Appendix H.2 shows an experiment verifying our classification bound (Theorem 4.3) and Appendix H.3 shows an experiment verifying our regression bound (Theorem 4.8 and 4.9). The experiment details are in Appendix I.
6.1 Swiss Roll Experiment
We first perform an experiment in a vertically separated Swiss Roll data distribution, see Figure 7a<sup>5</sup>. This example, similar to that in Section 5, demonstrates that a  $C_2$ -invariant model effectively "flattens" the  $z$ -dimension of the data so it must learn the decision boundary between two spirals (Figure 7b), whereas the non-equivariant model only needs to learn a horizontal plane to separate the classes, a significantly easier task. Besides the extrinsic data distribution, we consider two other data distributions shown in Figure 7c and Figure 7d, where a  $C_2$ -invariant model will observe incorrect and correct equivariance due to the mismatched and matched data labels in the two  $z$  planes.
We combine data from all three distributions in various proportions to test the performance of a  $z$ -invariant network (INV) with a baseline unconstrained network (MLP). Let  $c$  be the correct ratio, the proportion of data from the correct distribution. Define the incorrect ratio  $i$  and extrinsic ratio  $e$  similarly. We consider all  $c, i, e$  that are multiples of 0.125 such that  $c + i + e = 1$ . Figure 7ef shows some example data distributions. Relative to INV, this mixed data distribution has partial correct, incorrect, and extrinsic equivariance, which is not fully captured in prior work [58]. Based on Proposition 4.4, we have  $k(Gx) = 0.5$  for  $x$  drawn from the incorrect distribution, and  $k(Gx) = 0$  otherwise. Since the data is evenly distributed, we can calculate the error lower bound  $\mathrm{err}(h) \geq 0.5i$ .
4Notice that the four additional points in Figure 6b compared with (a) are not in the domain, they are created through applying the transformation rule of  $\mathcal{F}_E$  onto the domain.
For visualization, we show a simpler version of the data distribution. See Appendix H.1 for the actual one.
(a) Swiss Roll Data
(b)  $C_2$  -Invariant View
(c) Incorrect
(d) Correct
(e)  $c = .5, i = .5, e = 0$
(f)  $c = .5, i = 0, e = .5$
Results. Figure 8a shows the test success rate of INV compared with MLP when  $e$  and  $c$  vary with  $i = 0$ . When  $e$  increases, the performance of INV decreases while the performance of MLP shows an inverse trend, demonstrating that extrinsic equivariance is harmful in this experiment. Figure 8b shows the performance of INV and MLP when  $c$  and  $i$  vary while  $e = 0$ . The green line shows the upper bound of the test success rate  $(1 - 0.5i)$ . The experimental result matches our theoretical analysis quite closely. Notice that when  $c$  increases, there is a bigger gap between the performance of the network and its theoretical upper bound, since classification in the correct distribution is a harder task. Appendix H.1 shows the complete results of this experiment.
Figure 7: (a) (b) The Swiss Roll data distribution that leads to harmful extrinsic equivariance. (c) (d) The correct and incorrect data distribution in the Swiss Roll experiment. Here the spirals overlap with mismatched and matched labels respectively. (e) (f) Data distribution example with different correct ratio  $(c)$ , incorrect ratio  $(i)$ , and extrinsic ratio  $(e)$  values.
(a)
Figure 8: Result of the Swiss Roll experiment. (a) test success rate of an invariant network (red) and an unconstrained MLP (blue) with different extrinsic and correct ratio when incorrect ratio is 0. (b) same as (a) with different correct and incorrect ratio when extrinsic ratio is 0. Averaged over 10 runs.
6.2 Digit Classification Experiment
In this experiment, we apply our theoretical analysis to a realistic digit classification task using both the printed digit dataset [16] and the MNIST handwritten digit dataset [15]. We compare a  $D_{4}$ -invariant network  $(D_{4})$  with an unconstrained CNN. In the printed digit classification,  $D_{4}$  exhibits incorrect equivariance for 6 and 9 under a  $\pi$  rotation. Using Theorem 4.3, we can calculate a lower bound of error for  $D_{4}$  at  $10\%$ . However, as shown in Table 1 (top), the experimental results indicate that the actual performance is slightly better than predicted by the theory. We hypothesize that this discrepancy arises because a rotated 9 differs slightly from a 6 in some fonts. We conduct a similar experiment using the MNIST handwritten digit dataset (Table 1 bottom), where  $D_{4}$  achieves even better performance in classifying 6 and 9. This improvement is likely due to the more distinguishable handwriting of these digits, although the performance still underperforms the CNN as incorrect equivariance persists. It is important to note that there is a significant decrease in performance for  $D_{4}$  when classifying 2/5 and 4/7 compared to the CNN. This is because a vertical flip results in incorrect equivariance when classifying handwritten 2/5, and a similar issue arises for 4/7 under a  $\pi /2$  rotation followed by a vertical flip (notice that Weiler and Cesa [60] make a similar observation). These experiments demonstrate that our theory is useful not only for calculating the performance bounds of an equivariant network beforehand, but also for explaining the suboptimal performance of an equivariant network, thereby potentially assisting in model selection.
6.3 Robotic Experiment
In this experiment, we evaluate our theory in behavior cloning in robotic manipulation. We first preform an experiment where the problem is a mixture of correct and incorrect equivariance for a  $D_{1}$ -equivariant policy network  $(D_{1})$  where the robot's action will flip when the state is flipped
Digit	Overall	0	1	2	3	4	5	6	7	8	9
Print CNN	96.62	99.81	98.27	97.06	96.12	98.04	92.9	94.93	97.85	93.65	96.13
Print D4	92.5	99.71	98.62	96.94	95.98	97.91	93.52	63.08	98.42	95.88	76.17
Print D4 Upper Bound	90	100	100	100	100	100	100	50	100	100	50
MNIST CNN	98.21	99.51	99.61	98.62	98.83	98.08	98.47	97.99	97.04	96.98	96.81
MNIST D4	96.15	98.93	99.21	91.84	98.28	95.49	95.04	93.71	95.67	97.73	95.34
Table 1:  ${\bar{D}}_{4}$  -invariant network compared with an unconstrained CNN in printed and MNIST handwritten digit classification tasks. Bold indicates that there is a  $> 1\%$  difference in two models.
Stacking with Correct Equi.
Probability of Stacking (correct equi.):  $c$
Sorting with Incorrect Equi.
Probability of Sorting (incorrect equi.):  $1 - c$
Sorting with Extrinsic Equi.
Figure 9: Left: an environment containing both correct (Stacking) and incorrect (Pushing) equivariance for a  $D_{1}$ -equivariant (horizontal flip) policy net. Right: an environment with harmful extrinsic equivariance for the same policy.
Figure 10: Result of robotic experiment with different correct ratio. Averaged over 4 runs.
horizontally. Specifically, the environment contains two possible tasks (Figure 9 left). Stacking requires the robot to stack a green triangle on top of a blue cube. Here flip equivariance is correct. Sorting requires the robot to push the red cube to the left and the yellow triangle to the right. Here flip equivariance is incorrect because the robot should not sort the objects in an opposite way when the state is flipped (in other words,  $D_{1}$  cannot distinguish left and right). We vary the probability  $c$  of the stacking task (correct ratio) in the task distribution, and compare the performance of  $D_{1}$  versus a standard CNN policy. If we view the sorting task as a binary classification task, we can calculate an upper bound of performance for  $D_{1}$  using Theorem 4.3:  $0.5 + 0.5c$ . Figure 10 shows the result. Notice that the performance of  $D_{1}$  closely matches the theoretical upper bound, while the performance of CNN remains relatively stable for all Stacking-Sorting distributions.
We further evaluate  $D_{1}$  in a sorting task with harmful extrinsic equivariance. Here, the goal for the robot is the same as sorting above (push the red cube left and the yellow triangle right), however, the left and right sides of the workspace can now be differentiated by gray-scale colors. The shades of gray are discretized evenly into  $n$  bins, where the left side's color is randomly sampled from the odd-numbered bins, and the right side's color is randomly sampled from the even-numbered bins (Figure 9 right). The different color distributions of the left and right sides make  $D_{1}$  extrinsically equivariant, but it needs to learn the color distribution to distinguish left and right (while CNN can distinguish left and right directly). We set  $n = 10$ , and  $D_{1}$  achieves  $71.5 \pm 1.6\%$  test success rate, while CNN achieves  $99.5 \pm 0.5\%$ , demonstrating that the  $D_{1}$  extrinsic equivariance is harmful in this task. See Appendix I.5 for the details of the robot experiment.
7 Discussion
This paper presents a general theory for when the symmetry of the ground truth function and equivariant network are mismatched. We define pointwise correct, incorrect, and extrinsic equivariance, generalizing prior work [58] to include continuous mixtures of the three extremes. We prove error lower bounds for equivariant networks applied to asymmetric tasks including classification, invariant regression, and equivariant regression without the assumption of invariant data density. Our work discusses the potential disadvantage of extrinsic equivariance, and provides experiments that validate our theoretical analysis. The major limitation of this paper is that our theoretical lower bounds require domain knowledge like the density function over the domain. In future work, we will develop easy-to-apply model selection tools using our theory. Another future direction is theoretically understanding when extrinsic equivariance is helpful or harmful and analyzing the effect of extrinsic equivariance on the decision boundary of an equivariant network.
This work is supported in part by NSF 1724257, NSF 1724191, NSF 1763878, NSF 1750649, NSF 2107256, NSF 2134178, NSF 2312171, and NASA 80NSSC19K1474.
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A Integrals on the Group
Fundamental Domains In this paper, we are interested in cases in which the group  $G$  is not necessarily discrete but may have positive dimension. We do not assume the fundamental domain has non-empty interior, and thus domain is a misnomer. In this case the conjugates of the fundamental domain  $gF$  have measure 0 and the condition that their intersection have measure 0 is vacuous. Instead we assume a stronger condition, that the union of all pairwise intersections  $\bigcup_{g_1 \neq g_2} (g_1F \cap g_2F)$  has measure 0. We also require that  $F$  and the orbits  $Gx$  are differentiable manifolds such that integrals over  $X$  may be evaluated  $\int_X f(x) dx = \int_F \int_{Gy} f(z) dz dy$  similar to Equation 8 from [18].
Reparameterization Consider the integral
$$
\int_ {G x} f (z) d z. \tag {7}
$$
Denote the identification of the orbit  $Gx$  and coset space  $G / G_x$  with respect to the stabilizer  $G_x = \{g : gx = x\}$  by  $a_x \colon G / G_x \to Gx$ . Then the integral can be written
$$
\int_ {G / G _ {x}} f (\bar {g} x) \left| \frac {\partial a _ {x} (\bar {g})}{\partial \bar {g}} \right| d \bar {g}.
$$
We can also lift the integral to  $G$  itself
$$
\begin{array}{l} \int_ {G / G _ {x}} f (\bar {g} x) \left| \frac {\partial a _ {x} (\bar {g})}{\partial \bar {g}} \right| d \bar {g} = \left(\int_ {G _ {x}} d h\right) ^ {- 1} \left(\int_ {G _ {x}} d h\right) \int_ {G / G _ {x}} f (\bar {g} x) \left| \frac {\partial a _ {x} (\bar {g})}{\partial \bar {g}} \right| d \bar {g} \\ = \left(\int_ {G _ {x}} d h\right) ^ {- 1} \int_ {G / G _ {x}} \int_ {G _ {x}} f (\bar {g} h x) \left| \frac {\partial a _ {x} (\bar {g})}{\partial \bar {g}} \right| d h d \bar {g} \\ = \left(\int_ {G _ {x}} d h\right) ^ {- 1} \int_ {G} f (g x) \left| \frac {\partial a _ {x} (\bar {g})}{\partial \bar {g}} \right| d g. \\ \end{array}
$$
Define  $\alpha(g, x) = \left( \int_{G_x} dh \right)^{-1} \left| \frac{\partial a_x(\bar{g})}{\partial \bar{g}} \right|$ . Then
$$
\int_ {G x} f (z) d z = \int_ {G} f (g x) \alpha (g, x) d g.
$$
B Iterated Integral