| Approach | Training Metrics | Validation Metrics |
| Codebook Usage (↑) | Rec. Loss (↓) | Quantization Error (↓) | Rec. Loss (↓) | r-FID (↓) | r-IS (↑) |
| Codebook Lookup: Euclidean & Latent Shape: 32 × 32 × 32 & Codebook Size: 1024 |
| VQ-VAE | 100% | 0.107 | 5.9e-3 | 0.115 | 106.1 | 11.7 |
| VQ-VAE w/ Rotation Trick | 97% | 0.116 | 5.1e-4 | 0.122 | 85.7 | 17.0 |
| Codebook Lookup: Cosine & Latent Shape: 32 × 32 × 32 & Codebook Size: 1024 |
| VQ-VAE | 75% | 0.107 | 2.9e-3 | 0.114 | 84.3 | 17.7 |
| VQ-VAE w/ Rotation Trick | 91% | 0.105 | 2.7e-3 | 0.111 | 82.9 | 18.1 |
| Codebook Lookup: Euclidean & Latent Shape: 64 × 64 × 3 & Codebook Size: 8192 |
| VQ-VAE | 100% | 0.028 | 1.0e-3 | 0.030 | 19.0 | 97.3 |
| Gumbel VQ-VAE | 39% | 0.054 | — | 0.058 | 28.6 | 74.9 |
| VQ-VAE w/ Hessian Approx. | 39% | 0.082 | 6.9e-5 | 0.112 | 35.6 | 65.1 |
| VQ-VAE w/ Exact Gradients | 84% | 0.050 | 2.0e-3 | 0.053 | 25.4 | 80.4 |
| VQ-VAE w/ Rotation Trick | 99% | 0.028 | 1.4e-4 | 0.030 | 16.5 | 106.3 |
| Codebook Lookup: Cosine & Latent Shape: 64 × 64 × 3 & Codebook Size: 8192 |
| VQ-VAE | 31% | 0.034 | 1.2e-4 | 0.038 | 26.0 | 77.8 |
| VQ-VAE w/ Hessian Approx. | 37% | 0.035 | 3.8e-5 | 0.037 | 29.0 | 71.5 |
| VQ-VAE w/ Exact Gradients | 38% | 0.035 | 3.6e-5 | 0.037 | 28.2 | 75.0 |
| VQ-VAE w/ Rotation Trick | 38% | 0.033 | 9.6e-5 | 0.035 | 24.2 | 83.9 |
+
+angular distance to the chosen codebook vector are "pushed" to other, possibly unused, codebook regions by outwards-pointing gradients, thereby increasing codebook utilization. Concurrent with this effect, center-pointing gradients will "pull" points loosely clustered around the codebook vector closer together, locking on to the chosen codebook vector and reducing quantization error.
+
+# 4.4 FURTHER ANALYSIS
+
+The Appendix contains several supplementary analyses. Appendix A.2 compares the rotation trick with the STE for a non-convex synthetic example; Appendix A.4 looks at the behavior far away from the origin; and Appendix A.8 analyzes the effect of using a reflection rather than a rotation. Finally, Appendix A.9 examines scaling the gradient's norm by $\frac{\|q\|}{\|e\|}$ and explores alternatives.
+
+# 5 EXPERIMENTS
+
+In Section 4.3, we showed the rotation trick enables behavior that would increase codebook utilization and reduce quantization error by changing how points within the same Voronoi region are updated. However, the extent to which these changes will affect applications is unclear. In this section, we evaluate the effect of the rotation trick across many different VQ-VAE paradigms.
+
+We begin with image reconstruction: training a VQ-VAE with the reconstruction objective of Van Den Oord et al. (2017) and later extend our evaluation to the more complex VQGANs (Esser et al., 2021), the VQGANs designed for latent diffusion (Rombach et al., 2022), and then the ViT-VQGAN (Yu et al., 2021). Finally, we evaluate VQ-VAE reconstructions on videos using a TimeSformer (Bertasius et al., 2021) encoder and decoder. Due to space constraints, the video results are presented in Appendix A.1. In total, our empirical analysis spans 11 different VQ-VAE configurations. For all experiments, aside from handling $\frac{\partial q}{\partial e}$ differently, the models, hyperparameters, and training settings are identical and described in Appendix A.10.
+
+# 5.1 VQ-VAE EVALUATION
+
+We begin with a straightforward evaluation: training a VQ-VAE to reconstruct examples from ImageNet (Deng et al., 2009). Following Van Den Oord et al. (2017), our training objective is a linear combination of the reconstruction, codebook, and commitment loss:
+
+$$
+\mathcal {L} = \left\| x - \tilde {x} \right\| _ {2} ^ {2} + \left\| s g (e) - q \right\| _ {2} ^ {2} + \beta \left\| e - s g (q) \right\| _ {2} ^ {2}
+$$
+
+where $\beta$ is a hyperparameter scaling constant. Following convention, we drop the codebook loss term from the objective and instead use an exponential moving average to update the codebook vectors.
+
+Evaluation Settings. For $256 \times 256 \times 3$ input images, we evaluate two different settings: (1) compressing to a latent space of dimension $32 \times 32 \times 32$ with a codebook size of 1024 following Yu et al. (2021) and (2) compressing to $64 \times 64 \times 3$ with a codebook size of 8192 following Rombach et al. (2022). In both settings, we compare with a Euclidean and cosine similarity codebook lookup.
+
+Table 2: Results for VQGAN designed for autoregressive generation as implemented in https://github.com/ CompVis/taming-transformers. Experiments on ImageNet and the combined dataset FFHQ (Karras et al., 2019) and CelebA-HQ (Karras, 2017) use a latent bottleneck of dimension ${16} \times {16} \times {256}$ with 1024 codebook vectors.
+
+ | TimeSformer-VQGAN Settings |
| BAIR Robot Pushing | UCF101 Action Recognition |
| Input size | 16 × 64 × 64 × 3 | 16 × 128 × 128 × 3 |
| Patch size | 2 | 4 |
| Encoder / Decoder Hidden Dim | 256 | 256 |
| Encoder / Decoder MLP Dim | 768 | 768 |
| Encoder / Decoder Hidden Depth | 8 | 8 |
| Encoder / Decoder Hidden Num Heads | 4 | 4 |
| Codebook Dimension | 32 | 32 |
| Codebook Size | 1024 | 2048 |
| Codebook Loss Coefficient | 1.0 | 1.0 |
| Log Laplace loss Coefficient | 0.0 | 0.0 |
| Log Gaussian Coefficient | 1.0 | 1.0 |
| Perceptual loss Coefficient | 0.1 | 0.1 |
| Adversarial loss Coefficient | 0.1 | 0.1 |
| [Effective] Batch size | 24 | 20 |
| Learning rate | 1 × 10-4 | 4.5 × 10-6 |
| Weight Decay | 1 × 10-4 | 1 × 10-4 |
| Training epochs | 30 | 3 |
+
+We similarly fix constant gradient vectors for each partition, and apply them to the pre-quantized points after transformation by the STE, i.e. simply moving the gradient to each pre-quantized point in the quantized region, or by the rotation trick, i.e. rotating the gradient based on the angle between the pre-quantized point and closest codebook vector and rescaling appropriately. We multiply the gradient by a small constant—the learning rate—and then apply the gradient to each pre-quantized point. We repeat the above 25 times, at each point re-computing the angle and magnitude between the pre-quantized point and the codebook vector for the rotation trick update. For simplicity, we do not update the codebook vectors themselves or recompute codebook regions throughout the numerical simulation.
+
+# A.11 COMPARISON WITHIN GENERATIVE MODELING APPLICATIONS
+
+Absent from our work is an analysis on the effect of VQ-VAEs trained with the rotation trick on down-stream generative modeling applications. We see this comparison as outside the scope of this
+
+work and do not claim that improving reconstruction metrics, codebook usage, or quantization error in "Stage 1" VQ-VAE training will lead to improvements in "Stage 2" generative modeling applications.
+
+While poor reconstruction performance will clearly lead to poor generative modeling, recent work (Yu et al., 2023) suggests that—at least for autoregressive modeling of codebook sequences with MaskGit (Chang et al., 2022)—the connection between VQ-VAE reconstruction performance and downstream generative modeling performance is non-linear. Specifically, increasing the size of the codebook past a certain amount will improve VQ-VAE reconstruction performance but make downstream likelihood-based geneative modeling of codebook vectors more difficult.
+
+We believe this nuance may extend beyond MaskGit, and that the desiderata for likelihood-based generative models will likely be different than that for score-based generative models like diffusion. It is even possible that different preferences appear within the same class. For example, left-to-right autoregressive modeling of codebook elements with Transformers (Vaswani, 2017) may exhibit different preferences for Stage 1 VQ-VAE models than those of MaskGit.
+
+These topics deserve a deep, and rich, analysis that we would find difficult to include within this work as our focus is on propagating gradients through vector quantization layers. As a result, we entrust the exploration of these questions to future work.
+
+# A.12 GRADIENT ESTIMATORS AS PARALLEL TRANSPORT
+
+In this section, we analyze the STE and the rotation trick through the lens of differential geometry, specifically as the parallel transport of the gradient $\nabla_q\mathcal{L}$ vector from the codeword $q$ to the encoder output $e$ . For this analysis in this section, we only consider the rotational component $R_{\theta}$ of the rotation trick, not the rescaling by $\frac{\|q\|}{\|e\|}$ .
+
+# A.12.1 BACKGROUND ON HYPERSPHERICAL COORDINATES
+
+Hyperspherical coordinate systems are ubiquitous in applications of math and physics, where certain formulas become greatly simplified by parameterizing the location of points by the radius and angles to coordinate axes. An familiar instantiation of the hyperspherical coordinate system may be polar coordinates with radial component $r$ and polar angle $\theta$ :
+
+$$
+x = r \cos \theta
+$$
+
+$$
+y = r \sin \theta
+$$
+
+or the instantiation of the hyperspherical coordinate system for three dimensions, otherwise known as spherical coordinates, with radial component $r$ , polar angle $\theta$ and azimuthal angle $\phi$ :
+
+$$
+x = r \cos \theta
+$$
+
+$$
+y = r \sin \theta \cos \phi
+$$
+
+$$
+z = r \sin \theta \sin \phi
+$$
+
+More generally, hyperspherical coordinates are composed by a radial coordinate $r$ and $d - 1$ angular coordinates $\theta_{1},\ldots ,\theta_{d - 1}$ where $\theta_{1},\ldots \theta_{d - 2}$ are supported over $[0,\pi ]$ while $\theta_{d - 1}$ ranges from $[0,2\pi ]$ . We outline one common conversion from Cartesian coordinates to hyperspherical coordinates below, and other conversions are equivalent up to permutation of the coordinate axes:
+
+$$
+x _ {1} = r \cos (\theta_ {1})
+$$
+
+$$
+x _ {2} = r \sin (\theta_ {1}) \cos (\theta_ {2})
+$$
+
+$$
+x _ {3} = r \sin (\theta_ {1}) \sin (\theta_ {2}) \cos (\theta_ {3})
+$$
+
+\*
+
+$$
+x _ {d - 1} = r \sin (\theta_ {1}) \dots \sin (\theta_ {d - 2}) \cos (\theta_ {d - 1})
+$$
+
+$$
+x _ {d} = r \sin (\theta_ {1}) \dots \sin (\theta_ {d - 2}) \sin (\theta_ {d - 1})
+$$
+
+
+Figure 19: Visualization of basis vectors at different points under Cartesian (left) and spherical (right) coordinate systems. Notice that the Cartesian basis vectors do not change from point-to-point; however, the spherical basis vectors change in both direction and magnitude. Even at the same radius, the $\frac{\partial}{\partial\phi}$ coordinate changes based on the azimuth angle $\theta$ because the same infinitesimal change in $\phi$ will result in a longer (or smaller) change in arclength depending on the radius of the circle at latitude $\theta$ .
+
+
+
+and the reverse transform from Cartesian coordinates to hyperspherical coordinates:
+
+$$
+r = \sqrt {(x _ {1}) ^ {2} + (x _ {2}) ^ {2} + \ldots + (x _ {d}) ^ {2}}
+$$
+
+$$
+\theta_ {1} = \arctan 2 \left(\sqrt {\left(x _ {d}\right) ^ {2} + \dots + \left(x _ {2}\right) ^ {2}}, x _ {1}\right)
+$$
+
+$$
+\theta_ {2} = \arctan 2 \left(\sqrt {\left(x _ {d}\right) ^ {2} + \dots + \left(x _ {3}\right) ^ {2}}, x _ {2}\right)
+$$
+
+中
+
+$$
+\theta_ {d - 2} = \arctan 2 \left(\sqrt {\left(x _ {d}\right) ^ {2} + \left(x _ {d - 1}\right) ^ {2}}, x _ {d - 2}\right)
+$$
+
+$$
+\theta_ {d - 1} = \arctan 2 (\sqrt {(x _ {d}) ^ {2}}, x _ {d - 1})
+$$
+
+where $\arctan 2(x,y)$ returns the angle measurement in radians over the support $(-\pi ,\pi ]$ between between $x$ and $y$
+
+Unlike the Cartesian coordinate system, the hyperspherical basis vectors are not identical over the entire space; they change with position. For instance, moving outwards along $r$ will increase the length of $\frac{\partial}{\partial \theta^i}$ as an infinitesimal change in $\theta^i$ will now cover a larger arclength distance—i.e. the line segment traveled by changing the angle $\theta^i$ —than that same infinitesimal change with a smaller $r$ . This effect is visualized for three dimensions in Figure 19.
+
+At any given point in hyperspherical coordinates $\tilde{p}$ , the transformation from Cartesian basis vectors $\frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \ldots$ to hyperspherical basis vectors $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta^1}, \ldots$ can be computed with the multivariate chain rule:
+
+$$
+\frac {\partial}{\partial \theta^ {i}} = \sum_ {k = 1} ^ {d} \frac {\partial x ^ {k}}{\partial \theta^ {i}} \frac {\partial}{\partial x ^ {k}}
+$$
+
+where $\frac{\partial x^i}{\partial\theta^i}$ can be computed from the coordinate transform functions, i.e. $x_{1} = r\cos (\theta_{1})$ . It is typical to express these relationships in a matrix that transforms an arbitrary vector $v$ in Cartesian coordinates
+
+at point $p$ to its counterpart in hyperspherical coordinates $\tilde{v}$ at $\tilde{p}$ :
+
+$$
+\left[ \begin{array}{c c c c} \frac {\partial}{\partial x ^ {1}} & \frac {\partial}{\partial x ^ {2}} & \dots & \frac {\partial}{\partial x ^ {d}} \end{array} \right] \underbrace {\left[ \begin{array}{c c c c} \frac {\partial x ^ {1}}{\partial r} & \frac {\partial x ^ {1}}{\partial \theta^ {1}} & \dots & \frac {\partial x ^ {1}}{\partial \theta^ {d - 1}} \\ \frac {\partial x ^ {2}}{\partial r} & \frac {\partial x ^ {2}}{\partial \theta^ {1}} & \dots & \frac {\partial x ^ {2}}{\partial \theta^ {d - 1}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac {\partial x ^ {d}}{\partial r} & \frac {\partial x ^ {d}}{\partial \theta^ {1}} & \dots & \frac {\partial x ^ {d}}{\partial \theta^ {d - 1}} \end{array} \right]} _ {\text {T h e J a c o b i a n J}} = \left[ \begin{array}{c c c c} \frac {\partial}{\partial r} & \frac {\partial}{\partial \theta^ {1}} & \dots & \frac {\partial}{\partial \theta^ {d - 1}} \end{array} \right]
+$$
+
+As illustrated in Figure 19, $J$ does not necessarily have determinant equal to one and changes as a function of position, so the norms of the basis vectors spanning the hyperspherical tangent space change based on position. More generally, this notion of distance is captured by the line element: the length of a line segment resulting from an infinitesimal change along the coordinate axes. The Cartesian line element is given by:
+
+$$
+d s ^ {2} = (d x ^ {1}) ^ {2} + (d x ^ {2}) ^ {2} + \dots + (d x ^ {d}) ^ {2}
+$$
+
+while the hyperspherical line element is:
+
+$$
+d s ^ {2} = d r ^ {2} + r ^ {2} (d \theta^ {1}) ^ {2} + r ^ {2} \sin^ {2} \theta_ {1} (d \theta^ {1}) ^ {2} + r ^ {2} \left[ \prod_ {i = 2} ^ {d - 1} \sin^ {2} \theta_ {i} \right] (d \theta^ {d - 1}) ^ {2}
+$$
+
+which reflects that distance traveled by small changes in the hyperspherical coordinates "increases" with increasing radius and "decreases" with distance from the equator. To ensure that the norm of the basis vectors does not change during conversion, it is common to renormalize hyperspherical basis vectors to have unit norm for all points. However, a notion of norm is not defined a priori for hyperspherical vectors; the metric tensor imposed on this space defines the inner product which in turn defines a sense of arclength.
+
+Using the induced metric from Cartesian coordinates, we can inherit the inner product from Cartesian coordinates on the hyperspherical coordinate system by expressing hyperspherical basis vectors as a linear combination of Cartesian basis vectors and then computing the norm of this resulting vector in the Cartesian tangent space:
+
+$$
+\begin{array}{l} \| \frac {\partial}{\partial \theta^ {i}} \| = \sqrt {\left\langle \frac {\partial}{\partial \theta^ {i}} , \frac {\partial}{\partial \theta^ {i}} \right\rangle} \\ = \sqrt {\left[ \sum_ {k = 1} ^ {d} \frac {\partial x ^ {k}}{\partial \theta^ {i}} \frac {\partial}{\partial x ^ {k}} \right] \cdot \left[ \sum_ {j = 1} ^ {d} \frac {\partial x ^ {j}}{\partial \theta^ {i}} \frac {\partial}{\partial x ^ {j}} \right]} \\ = \sqrt {\sum_ {k = 1} ^ {d} \frac {\partial x ^ {k}}{\partial \theta^ {i}} \frac {\partial x ^ {k}}{\partial \theta^ {i}} \left[ \frac {\partial}{\partial x ^ {k}} \cdot \frac {\partial}{\partial x ^ {k}} \right]} \\ = \sqrt {\sum_ {k = 1} ^ {d} \left(\frac {\partial x ^ {k}}{\partial \theta^ {i}}\right) ^ {2}} \\ \end{array}
+$$
+
+The first fundamental form gives us the normalization constants:
+
+$$
+\mathcal {I} = \left[ \begin{array}{l l l l l} 1 ^ {2} & 0 & 0 & \dots & 0 \\ 0 & r ^ {2} & 0 & \dots & 0 \\ 0 & 0 & r ^ {2} \sin^ {2} \theta_ {1} & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & r ^ {2} \prod_ {i = 1} ^ {d - 1} s i n ^ {2} \theta_ {i} \end{array} \right]
+$$
+
+as the diagonal represents the inner product $\langle \frac{\partial}{\partial\theta^i},\frac{\partial}{\partial\theta^i}\rangle$ , and we would like to renormalize each basis vector to have unit norm: $\| \frac{\partial}{\partial\theta^i}\| = \sqrt{\langle\frac{\partial}{\partial\theta^i},\frac{\partial}{\partial\theta^i}\rangle}$ . Therefore, our normalized hyperspherical basis vectors $\frac{\partial}{\partial\hat{r}},\frac{\partial}{\partial\hat{\theta}_1},\ldots$ become:
+
+$$
+\frac {\partial}{\partial \hat {r}} = \frac {\partial}{\partial r}
+$$
+
+$$
+\frac {\partial}{\partial \hat {\theta} _ {i}} = (\mathcal {I} _ {i i}) ^ {- \frac {1}{2}} \frac {\partial}{\partial \theta_ {i}}
+$$
+
+Using our convention from earlier, we can now compute the transformation from Cartesian basis vectors to normalized hyperspherical basis vectors:
+
+$$
+\frac {\partial}{\partial \hat {\theta} ^ {i}} = (\mathcal {I} _ {i i}) ^ {- \frac {1}{2}} \sum_ {k = 1} ^ {d} \frac {\partial x ^ {k}}{\partial \theta^ {i}} \frac {\partial}{\partial x ^ {k}}
+$$
+
+to compose the normalized "Jacobian" $\hat{J}$ :
+
+$$
+\left[ \begin{array}{c c c c} \frac {\partial}{\partial x ^ {1}} & \frac {\partial}{\partial x ^ {2}} & \dots & \frac {\partial}{\partial x ^ {d}} \end{array} \right] \underbrace {\left[ \begin{array}{c c c c} \frac {\partial x ^ {1}}{\partial \hat {r}} & \frac {\partial x ^ {1}}{\partial \hat {\theta} _ {1}} & \dots & \frac {\partial x ^ {1}}{\partial \hat {\theta} _ {d - 1}} \\ \frac {\partial x ^ {2}}{\partial \hat {r}} & \frac {\partial x ^ {2}}{\partial \hat {\theta} _ {1}} & \dots & \frac {\partial x ^ {2}}{\partial \hat {\theta} _ {d - 1}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac {\partial x ^ {d}}{\partial \hat {r}} & \frac {\partial x ^ {d}}{\partial \hat {\theta} _ {1}} & \dots & \frac {\partial x ^ {d}}{\partial \hat {\theta} _ {d - 1}} \end{array} \right]} _ {\hat {J} \in S O (d)} = \left[ \begin{array}{c c c c} \frac {\partial}{\partial \hat {r}} & \frac {\partial}{\partial \hat {\theta} ^ {1}} & \dots & \frac {\partial}{\partial \hat {\theta} ^ {d - 1}} \end{array} \right] \tag {1}
+$$
+
+Rescaling the hyperspherical basis vectors to have unit norm at all points causes the matrix $J$ to become the orthogonal matrix with determinant equal to one $\hat{J}$ . This set of $d \times d$ matrices belongs to the group $SO(d)$ , which represents the set of $d$ -dimensional rotations about the origin. Similarly, the backwards change-of-basis $\hat{J}^{-1} = \hat{J}^T$ converts vectors in hyperspherical coordinates to Cartesian coordinates.
+
+As a result, vectors from the tangent space at $p$ in Cartesian coordinates simply rotate to convert to the normalized tangent space at $\tilde{p}$ in hyperspherical coordinates. Specifically, for a point $\tilde{p} = (r, \theta_1, \theta_2, \dots, \theta_{d-1})$ and a vector $\tilde{v} = \tilde{c}_1 r + \tilde{c}_2 \theta^1 + \dots + \tilde{c}_d \theta^{d-1}$ , converting $v = c_1 x^1 + \dots + c_d x^d$ from Cartesian to hyperspherical coordinates is the transformation:
+
+$$
+\tilde {v} = \hat {J} ^ {T} v
+$$
+
+where $\hat{J}$ operates on vector $v$ —i.e. $\hat{J} v$ —by first rotating by angle $\tilde{c}_2$ in the $x^1 - x^2$ plane (i.e. the $\theta^1$ axis of rotation), then by angle $\tilde{c}_3$ in the $x^2 - x^3$ plane (i.e. the $\theta^2$ axis of rotation), so on and so forth until a final rotation by angle $\tilde{c}_d$ in the $x^{d-1} - x^d$ plane (i.e. the $\theta^{d-1}$ axis of rotation). Composing these rotations together leads to a rotation from $\tilde{p}_0 = (1, 0, 0, \dots, 0)$ to $\tilde{p}$ :
+
+$$
+\hat {J} v = (R _ {\tilde {p} _ {0} \rightarrow \tilde {p}}) v = (R _ {\theta_ {d}} ^ {x ^ {d - 1} - x ^ {d}} \dots R _ {\theta_ {2}} ^ {x ^ {2} - x ^ {3}} R _ {\theta_ {1}} ^ {x ^ {1} - x ^ {2}}) v
+$$
+
+$$
+\hat {J} ^ {- 1} v = \hat {J} ^ {T} v = (R _ {\tilde {p} _ {0} \rightarrow \tilde {p}}) ^ {T} v = (R _ {\theta_ {d}} ^ {x ^ {d - 1} - x ^ {d}} \dots R _ {\theta_ {2}} ^ {x ^ {2} - x ^ {3}} R _ {\theta_ {1}} ^ {x ^ {1} - x ^ {2}}) ^ {T} v = R _ {\tilde {p} \rightarrow \tilde {p} _ {0}} v
+$$
+
+where we define $R_{\tilde{a}\rightarrow \tilde{b}}$ to be the rotation from $\tilde{a}$ to $\tilde{b}$ as described above and $R_{\theta_i}^{x^i -x^j}$ to be the rotation by angle $\theta_{i}$ in the $x^{i} - x^{j}$ plane. Important for our later discussion on the rotation trick, this rotational characteristic causes moving a fixed vector along a curve in hyperspherical coordinates to rotate in Cartesian coordinates.
+
+Remark 4. Using the renormalized transformation in Equation (1), a constant vector field $\tilde{v}$ in hyperspherical coordinates corresponds to a rotated vector field in Cartesian coordinates.
+
+Proof. At Cartesian point $p$ and corresponding hyperspherical point $\tilde{p}$ :
+
+$$
+v _ {p} ^ {T} \left[ R _ {\theta_ {d}} ^ {x ^ {d - 1} - x ^ {d}} \dots R _ {\theta_ {2}} ^ {x ^ {2} - x ^ {3}} R _ {\theta_ {1}} ^ {x ^ {1} - x ^ {2}} \right] = \tilde {v} _ {\tilde {p}} ^ {T}
+$$
+
+$$
+\left[ R _ {\theta_ {d}} ^ {x ^ {d - 1} - x ^ {d}} \dots R _ {\theta_ {2}} ^ {x ^ {2} - x ^ {3}} R _ {\theta_ {1}} ^ {x ^ {1} - x ^ {2}} \right] ^ {T} v _ {p} = \tilde {v} _ {\tilde {p}}
+$$
+
+$$
+\left[ R _ {\tilde {p} \rightarrow \tilde {p} _ {0}} \right] v _ {p} = \tilde {v} _ {\tilde {p}}
+$$
+
+so a constant vector field $\tilde{v}$ in hyperspherical coordinates will correspond to a cartesian vector field where each vector at point $p$ is rotated by the rotation that alights $\tilde{p}$ to $\tilde{p}_0$ .
+
+Another important characteristic relates to the metric tensor with normalized hyperspherical basis vectors. We can explicitly compute the induced metric in hyperspherical coordinates in terms of our renormalized basis vectors:
+
+$$
+\begin{array}{l} \hat {\mathcal {L}} = \left[ \begin{array}{c c c c c} \frac {\partial}{\partial \hat {r}} \cdot \frac {\partial}{\partial \hat {r}} & 0 & 0 & \dots & 0 \\ 0 & \frac {\partial}{\partial \hat {\theta} _ {1}} \cdot \frac {\partial}{\partial \hat {\theta} _ {1}} & 0 & \dots & 0 \\ 0 & 0 & \frac {\partial}{\partial \hat {\theta} _ {2}} \cdot \frac {\partial}{\partial \hat {\theta} _ {2}} & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \frac {\partial}{\partial \hat {\theta} _ {d - 1}} \cdot \frac {\partial}{\partial \hat {\theta} _ {d - 1}} \end{array} \right] \\ = \left[ \begin{array}{c c c c c} (\mathcal {I} _ {1 1}) ^ {- 1} \frac {\partial}{\partial r} \cdot \frac {\partial}{\partial r} & 0 & 0 & \dots & 0 \\ 0 & (\mathcal {I} _ {2 2}) ^ {- 1} \frac {\partial}{\partial \theta_ {1}} \cdot \frac {\partial}{\partial \theta_ {1}} & 0 & \dots & 0 \\ 0 & 0 & (\mathcal {I} _ {3 3}) ^ {- 1} \frac {\partial}{\partial \theta_ {2}} \cdot \frac {\partial}{\partial \theta_ {2}} & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & (\mathcal {I} _ {d d}) ^ {- 1} \frac {\partial}{\partial \theta_ {d - 1}} \cdot \frac {\partial}{\partial \theta_ {d - 1}} \end{array} \right] \\ = \left[ \begin{array}{c c c c c} (\mathcal {I} _ {1 1}) ^ {- 1} (\mathcal {I} _ {1 1}) & 0 & 0 & \ldots & 0 \\ 0 & (\mathcal {I} _ {2 2}) ^ {- 1} (\mathcal {I} _ {2 2}) & 0 & \ldots & 0 \\ 0 & 0 & (\mathcal {I} _ {3 3}) ^ {- 1} (\mathcal {I} _ {3 3}) & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & (\mathcal {I} _ {d d}) ^ {- 1} (\mathcal {I} _ {d d}) \end{array} \right] \\ = \left[ \begin{array}{c c c c c} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{array} \right] \tag {2} \\ \end{array}
+$$
+
+which yields the identity matrix. This is perhaps unsurprising: we normalize basis vectors so that $\frac{\partial}{\partial\hat{\theta}^i}\cdot \frac{\partial}{\partial\hat{\theta}^i} = 1$
+
+Another way to view the renormalized tangent plane transformation is as a change-of-basis in the Cartesian coordinate system, with the basis vectors spanning each tangent space in Cartesian coordinates rotated to align with the directions of hyperspherical basis vectors. Rotating the tangent space at each point in Cartesian coordinates does not change the Euclidean metric tensor— $R^T IR = I$ so it remains the identity. These two formulations are equivalent: renormalizing the basis vectors in the hyperspherical tangent space to have unit norm corresponds to rotating the basis vectors in the Cartesian coordinate system to align with the hyperspherical basis vectors at all points.
+
+# A.12.2 STE AS PARALLEL TRANSPORT
+
+From the description of the STE in Bengio et al. (2013), a gradient vector $\nabla_q\mathcal{L}$ is transported from $q$ to $e$ during the backwards pass in such a way that its direction and magnitude is preserved. Critically, the curve along which $\nabla_q\mathcal{L}$ is transported is not specified; the effect is to simply "copy-and-paste" the vector from $q$ to $e$ .
+
+To use the machinery of calculus, we assume that $\nabla_q\mathcal{L}$ is transported from $q$ to $e$ along any smooth curve $\gamma (t)$ running from $q$ to $e$ . Along this curve, we define the transport of $\nabla_q\mathcal{L}$ at position $\gamma (t)$ simply as $\nabla_q\mathcal{L}$ to emulate how the STE would move $\nabla_q\mathcal{L}$ from $q$ to $\gamma (t)$ . Therefore, the direction and magnitude of $\nabla_q\mathcal{L}$ does not change along the curve $\gamma (t)$ . An example of this transport is visualized in Figure 20, and in Remark 5, we show this formulation is equivalent to the parallel transport of $\nabla_q\mathcal{L}$ along any curve $\gamma (t)$ from $q$ to $e$ with the Levi-Civita connection.
+
+Remark 5. The Straight Through Estimator (STE) is equivalent to the parallel transport of $\nabla_q\mathcal{L}$ along any curve connecting $q$ to $e$ with the identity metric tensor in Cartesian coordinates using the Levi-Civita connection.
+
+Proof. A vector field $v$ is parallel transported along a curve $\gamma(t)$ if the covariant derivative of $v$ in the direction of $\dot{\gamma}(t)$ is zero. Informally, the change in the vector field $v$ must exactly match how the
+
+
+Transport of $\nabla_q\mathcal{L}$ from q to e in Cartesian Coordinates in Hyperspherical Coordinates
+
+
+Cartesian Vector Field
+Figure 20: (top) Visualization of vector transport in Cartesian coordinates and renormalized hyperspherical coordinates along curves $\gamma_{1}(t)$ , $\gamma_{2}(t)$ and $\gamma_{3}(t)$ . Notice the hyperspherical basis changes from point to point. (bottom) Depiction of the transported vector in terms of the basis vectors $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ for Cartesian coordinates and $\frac{\partial}{\partial \hat{r}}$ and $\frac{\partial}{\partial \hat{\theta}}$ for hyperspherical coordinates. Notice how the components of $\frac{\partial}{\partial \hat{r}}$ and $\frac{\partial}{\partial \hat{\theta}}$ change for a constant vector field in the Cartesian tangent space.
+$\nabla_{\dot{\gamma} (t)}v = \vec{0}$ Parallel Transport Condition
+
+
+
+
+Hyperspherical Vector Field for $\gamma_3(t)$
+
+basis vectors of the tangent plane change along $\gamma (t)$ to remain "parallel" along the curve:
+
+Using the identity metric tensor:
+
+$$
+g _ {i j} = \delta_ {i j} = \left\{ \begin{array}{l} 0 \text {i f} i \neq j \\ 1 \text {i f} i = j \end{array} \right.
+$$
+
+with the Levi-Civita connection will result in all zero Christoffel symbols:
+
+$$
+\Gamma_ {i j} ^ {m} = \frac {1}{2} g ^ {m k} \left(\frac {\partial g _ {j k}}{\partial x ^ {i}} + \frac {\partial g _ {i k}}{\partial x ^ {j}} - \frac {\partial g _ {i j}}{\partial x ^ {k}}\right) = 0
+$$
+
+where $g^{mk}$ is the $m, k$ entry of inverse metric tensor. Computing the covariant derivative for a general curve $\gamma(t)$ :
+
+$$
+\begin{array}{l} \nabla_ {\dot {\gamma} (t)} v = \nabla_ {\dot {\gamma} _ {1} e _ {1} + \dot {\gamma} _ {2} e _ {2} + \dots + \dot {\gamma} _ {d} e _ {d}} v \\ = \sum_ {i = 1} ^ {d} \dot {\gamma} _ {i} \nabla_ {e _ {i}} v \\ = \sum_ {i = 1} ^ {d} \dot {\gamma} _ {i} \frac {\partial}{\partial x ^ {i}} (v) \\ = \sum_ {i = 1} ^ {d} \underbrace {\dot {\gamma} _ {i} \frac {\partial}{\partial x ^ {i}} \left(v _ {1} e _ {1} + v _ {2} e _ {2} + \dots + v _ {d} e _ {d}\right)} _ {\text {m u s t b e e q u a l t o 0 f o r p a r a l l e l} \\ \end{array}
+$$
+
+Considering the $i^{th}$ term in this summation:
+
+$$
+\begin{array}{l} 0 = \left(\nabla_ {\dot {\gamma} (t)} v\right) ^ {i} = \dot {\gamma} _ {i} \frac {\partial}{\partial x ^ {i}} \left(v _ {1} e _ {1} + v _ {2} e _ {2} + \dots + v _ {d} e _ {d}\right) \\ = \dot {\gamma} _ {i} \frac {\partial}{\partial x ^ {i}} \left[ v ^ {k} e _ {k} \right] \\ = \dot {\gamma} _ {i} \left[ \frac {\partial v ^ {k}}{\partial x ^ {i}} e _ {k} + v ^ {k} \frac {\partial e _ {k}}{\partial x ^ {i}} \right] \\ = \dot {\gamma} _ {i} \left[ \frac {\partial v ^ {k}}{\partial x ^ {i}} e _ {k} + v ^ {k} \Gamma_ {i k} ^ {m} e _ {m} \right] \\ = \dot {\gamma} _ {i} \left[ \frac {\partial v ^ {k}}{\partial x ^ {i}} e _ {k} \right] \\ \end{array}
+$$
+
+For this equation to hold for an arbitrary $\gamma(t)$ , $\frac{\partial v_k}{\partial x^i} = 0$ for $1 \leq k, i \leq d$ . Therefore, $v_k$ must be a constant, and vector fields along curves must be constant to satisfy the parallel transport criteria.
+
+Pulling this back to the STE, holding $\nabla_q\mathcal{L}$ constant along the curve $\gamma (t)$ from $q$ to $e$ results in a constant vector field along $\gamma (t)$ . The covariant derivative of this vector field is zero, and therefore the STE parallel transports $\nabla_q\mathcal{L}$ from $q$ to $e$ .
+
+# A.12.3 THE ROTATION TRICK AS PARALLEL TRANSPORT
+
+In this section, we analyze the rotation trick through the lens of geometry. As in Appendix A.12.2, we extend the rotation trick to any smooth curve $\gamma(t)$ connecting $q$ to $e$ and define the transport of $\nabla_q\mathcal{L}$ at $\gamma(t)$ as the rotation trick applied to move $\nabla_q\mathcal{L}$ from $q$ to $\gamma(t)$ . This definition allows us to use the structure of calculus, without imposing any prohibitive restrictions on the path taken from $q$ to $e$ .
+
+To build visual intuition, Figure 21 illustrates how the rotation trick transforms an initial vector along three different curves $\gamma_{1},\gamma_{2},\gamma_{3}$ in both Cartesian coordinates and hyperspherical coordinates with normalized basis vectors. In Cartesian coordinates, the rotation trick changes the components of the basis vectors during transport to follow a rotation; however in normalized hyperspherical coordinates, the components of this vector during transport are constant because the basis vectors themselves rotate.
+
+Remark 6. The rotation trick is equivalent to the parallel transport of $\nabla_q\mathcal{L}$ along any curve connecting $q$ to $e$ with the induced metric in hyperspherical coordinates with the normalized transformation described in Equation (1) using the Levi-Civita connection.
+
+Proof. From Equation (2), the metric tensor in hyperspherical coordinates with normalized basis vectors—equivalently, the cartesian coordinate system with each tangent space rotated to align with the hyperspherical frame at every point—is the identity. Therefore, using the Levi-Civita connection
+
+
+Transport of $\nabla_{\phi}$ in Hyperspherical Coordinates
+
+
+Hyperspherical Vector Field
+Figure 21: (top) Visualization of vector transport in hyperspherical coordinates with normalized basis vectors and Cartesian coordinates along curves $\gamma_{1}(t),\gamma_{2}(t)$ and $\gamma_3(t)$ . The vectors along each curve in hyperspherical coordinates rotate to stay constant with respect to the natural rotation of the basis vectors. This same rotation in Cartesian coordinates yields a non-constant vector as the Cartesian basis vectors do not change from point to point. (bottom) Depiction of the transported vector in terms of the basis vectors $\frac{\partial}{\partial\vec{r}}$ and $\frac{\partial}{\partial\vec{\theta}}$ for hyperspherical coordinates and $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ for Cartesian coordinates. In the former case, the transported vector remains constant with respect to the normalized basis vectors, while in Cartesian coordinates, the components change along $\gamma_3(t)$ .
+
+
+in Cartesian Coordinates
+
+
+Cartesian Vector Field
+
+leads to zero Christoffel symbols, and the parallel transport of a vector along any curve keeps the vector constant.
+
+We define $T_{p}C$ as the tangent space of the Cartesian coordinate system at point $p$ and $T_{\tilde{p}}H$ as the tangent space of the hyperspherical coordinate system with normalized basis vectors at point $\tilde{p}$ . It remains to show that for a vector $\nabla_q\mathcal{L}\in T_qC$ and corresponding $\nabla_{\tilde{q}}\tilde{\mathcal{L}}\in T_{\tilde{q}}H$ , the transformation of $\nabla_{\tilde{q}}\tilde{\mathcal{L}}\in T_{\tilde{e}}H$ to $T_{e}C$ will yield $R_{q\to e}\nabla_q\mathcal{L}$ where $R_{q\rightarrow e}$ is the rotation trick's transformation, i.e. the rotation that rotates $q$ to $e$ .
+
+For a vector $\nabla_{\tilde{q}}\tilde{\mathcal{L}}$ in hyperspherical coordinates at point $\tilde{q} = (1,\theta_1,\theta_2,\dots,\theta_{d - 1})$ and using the normalized change-of-basis in Equation (1), the corresponding vector $\nabla_q\mathcal{L}$ in Cartesian coordinates is:
+
+$$
+\begin{array}{l} \nabla_ {q} \mathcal {L} ^ {T} = \nabla_ {\tilde {q}} \tilde {\mathcal {L}} ^ {T} \left[ \hat {J} _ {\tilde {q}} ^ {- 1} \right] \\ \nabla_ {q} \mathcal {L} = \left[ \hat {J} _ {\tilde {q}} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ = \left[ R _ {\tilde {p} _ {0} \rightarrow \tilde {q}} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ = \left[ R _ {\theta_ {d - 1}} R _ {\theta_ {d - 2}} \dots R _ {\theta_ {1}} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ \end{array}
+$$
+
+and the corresponding vector $\nabla_{\tilde{q}}\tilde{\mathcal{L}}$ at point $\tilde{e}$ is:
+
+$$
+\begin{array}{l} \nabla_ {e} \mathcal {L} ^ {T} = \nabla_ {\tilde {q}} \tilde {\mathcal {L}} ^ {T} \left[ \hat {J} _ {e} ^ {- 1} \right] \\ \nabla_ {e} \mathcal {L} = \left[ \hat {J} _ {e} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ = \left[ R _ {\tilde {p} _ {0} \rightarrow \tilde {e}} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ = \left[ R _ {\tilde {q} \rightarrow \tilde {e}} R _ {\tilde {p} _ {0} \rightarrow \tilde {q}} \right] \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \\ = R _ {\tilde {q} \rightarrow \tilde {e}} \left[ R _ {\tilde {p} _ {0} \rightarrow \tilde {q}} \nabla_ {\tilde {q}} \tilde {\mathcal {L}} \right] \\ = [ R _ {\tilde {q} \rightarrow \tilde {e}} ] \nabla_ {q} \mathcal {L} \\ \end{array}
+$$
+
+which is exactly how the rotation trick transforms the vector. Informally, "copy-and-pasting" the vector $\nabla_{\tilde{q}}\tilde{\mathcal{L}}$ from $\tilde{q}$ to $\tilde{e}$ in hyperspherical coordinates with normalized basis vectors corresponds to rotating $\nabla_q\mathcal{L}$ by the rotation that aligns $q$ to $e$ in Cartesian coordinates.
+
+In summary, we consider a geometry where the tangent space is spanned by unit norm basis vectors $\frac{\partial}{\partial r}, \frac{\partial}{\partial \hat{\theta}^1}, \dots, \frac{\partial}{\partial \hat{\theta}^{d-1}}$ that match the direction of the typical hyperspherical basis vectors $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta^1}, \dots, \frac{\partial}{\partial \hat{\theta}^{d-1}}$ . The induced metric tensor is the identity, so the parallel transport of a vector along any curve holds its components constant. Converting a vector $\nabla_q \mathcal{L}$ to this tangent space via the normalized transformation in Equation (1), parallel transporting the resulting vector from $\tilde{q}$ to $\tilde{e}$ , and then converting it back to Cartesian coordinates corresponds exactly to the rotation trick's transformation.
+
+This is a remarkably simple result; the rotation trick and the STE can be viewed as the same operation. Both parallel transport the gradient $\nabla_q\mathcal{L}$ from $q$ to $e$ in a path-independent manner with the Euclidean metric. The only difference is the coordinate system where parallel transport occurs. The STE employs the Cartesian coordinate system while the rotation trick uses the hyperspherical coordinate system with normalized basis vectors.
\ No newline at end of file