Thread (26 posts)

You're absolutely right to question this - I was imprecise. A critical point (where a ReLU toggles) doesn't necessarily affect all output coordinates.

Here's what actually happens:

When a neuron in some hidden layer toggles at position $\lambda^*$ along the interpolation path, this creates a critical point at that specific $\lambda^*$ value. However, whether this creates a visible corner in each output coordinate depends on the weights connecting that neuron to each output.

For output coordinate $j$:

• If there's a non-zero path of weights from the toggling neuron to output $j$, then $f_j(\mu(\lambda))$ will have a corner at $\lambda^*$
• If all paths from that neuron to output $j$ have zero total weight, then $f_j(\mu(\lambda))$ remains smooth at $\lambda^*$

So the critical point occurs at the same location $\lambda^*$ for all outputs (since it's the same neuron toggling), but only manifests as a corner in outputs that are actually influenced by that neuron.

The paper's approach still works because:

1. Most neurons in a trained network influence at least some outputs
2. By analyzing multiple random interpolation paths, they collect critical points for different neurons
3. Even if some outputs don't show corners for certain neurons, enough output coordinates will show corners to detect the critical points

This is why the paper can work with either single or multiple outputs - they just need at least one output coordinate that shows a corner to detect each critical point.

Let me prove this using the definitions from the paper.

Definitions:

• ReLU: $\sigma(x) = \max(x, 0)$
• Neuron $j$ in layer $i$: $\eta_j^{(i)}(x) = \sigma(A_j^{(i)}x + b_j^{(i)})$
• Network: $f = f_{r+1} \circ \sigma \circ \cdots \circ \sigma \circ f_2 \circ \sigma \circ f_1$

Proof:

Consider an interpolation path $\mu(\lambda) = x_1 + \lambda(x_2 - x_1)$ for $\lambda \in [0,1]$.

For neuron $j$ in layer $i$, its pre-activation value along this path is: $v_j^{(i)}(\lambda) = A_j^{(i)} \cdot \text{(input to layer i at } \mu(\lambda)) + b_j^{(i)}$

This pre-activation value is a linear function of $\lambda$ (since the input to layer $i$ depends linearly on $\lambda$ when all neurons maintain their states). The neuron becomes critical when $v_j^{(i)}(\lambda^*) = 0$ for some $\lambda^*$.

Now consider the network's output. Let's denote the $k$-th output coordinate as $f_k$. We can write:

$f_k(\mu(\lambda)) = \sum_{j=1}^{d_r} w_{kj}^{(r+1)} \cdot \eta_j^{(r)}(\cdots) + b_k^{(r+1)}$

where $w_{kj}^{(r+1)}$ are the weights in the final layer.

The key observation: When neuron $j$ in layer $i$ toggles at $\lambda = \lambda^*$:

• For $\lambda < \lambda^*$: $\eta_j^{(i)} = 0$ (if going from inactive to active)
• For $\lambda > \lambda^*$: $\eta_j^{(i)} = A_j^{(i)} \cdot \text{(input)} + b_j^{(i)}$

This change propagates through all subsequent layers. Every path from this neuron to any output coordinate will experience a change in its contribution at exactly $\lambda = \lambda^*$.

Specifically, the derivative $\frac{d f_k}{d\lambda}$ will have a discontinuity at $\lambda^*$ for every output coordinate $k$ that has any dependency on neuron $j$ in layer $i$. Since the network is fully connected, every neuron in layer $i$ influences every output coordinate (unless some weights happen to be exactly zero).

Therefore, when neuron $j$ toggles at $\lambda^*$:

1. The function $f_k(\mu(\lambda))$ experiences a corner (slope change) at $\lambda^*$ for all $k$
2. The location $\lambda^*$ is the same for all output coordinates
3. Only the magnitude of the slope change differs between outputs

This is why the paper can find critical points by analyzing any output coordinate - the $\lambda^*$ values where corners occur are identical across all outputs.