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Jacob Helwig
Jacob Helwig
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Proof of rotation equivariance ofcovariant convolution for a kernel function that is rotation invariantsymmetric in Fourier space

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Jacob Helwig
Jacob Helwig
  • 53
  • 5

Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1} \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right). \label{2}\tag{2} \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation \eqref{1}.

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g. $$

For this convolution to be defined, bothIf $f$ and $g$ must beare integrable, then the convolution is defined ([source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1} \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right). \label{2}\tag{2} \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation \eqref{1}.

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g. $$

For this convolution to be defined, both $f$ and $g$ must be integrable ([source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1} \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right). \label{2}\tag{2} \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation \eqref{1}.

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g. $$

If $f$ and $g$ are integrable, then the convolution is defined ([source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

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LSpice
LSpice
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Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right) \quad\quad\quad (1) \end{equation}\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1} \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right) \quad\quad\quad (2) \end{equation}\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right). \label{2}\tag{2} \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation (\eqref{1)}.

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g $$$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g. $$

For this convolution to be defined, both $f$ and $g$ must be integrable ([source][source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation (\eqref{1)} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right) \quad\quad\quad (1) \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right) \quad\quad\quad (2) \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation (1).

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g $$

For this convolution to be defined, both $f$ and $g$ must be integrable ([source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation (1) to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

Problem Statement

Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption I1). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption C) is an integrable function (assumption I2) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption S). Finally, let assumption(s) U be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:

\begin{equation} \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1} \end{equation}

From the convolution theorem, the following is equivalent:

\begin{equation} f*\left(\mathcal Tg\right)=\mathcal T\left(f*g\right). \label{2}\tag{2} \end{equation}

In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.

My question is to define U and show how these assumptions can be used to prove equation \eqref{1}.

Assumptions

I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.

Assumption I1 and I2

From the convolution theorem:

$$ \mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f*g. $$

For this convolution to be defined, both $f$ and $g$ must be integrable ([source]).

Assumption C

Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption S.

Assumption S

Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).

deleted 115 characters in body
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Jacob Helwig
Jacob Helwig
  • 53
  • 5
Source Link
Jacob Helwig
Jacob Helwig
  • 53
  • 5