I'm looking at some formulas involving matrices (in the context of machine learning, but I'm not sure it's relevant) and I came across $\odot$. What could this mean? The context is $M \odot N$, where $M$ is a matrix and $N$ might be a vector, or a matrix, or a scalar, it's a bit dense so it's hard to tell. I have reason to believe it may be the Hadamard product, is there anything else it could mean?

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    $\begingroup$ Could you give any more context? It could just mean $M$ acts on $v$ by multiplication. $\endgroup$ – Ian Coley May 30 '14 at 20:17
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    $\begingroup$ I think you might have meant "dot in a circle", rather than "circle in a dot", in your title. $\endgroup$ – Robert Lewis May 30 '14 at 20:21
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    $\begingroup$ I have seen it used to define simple "non-standard" operators when learning group theory so that there is no confusion with $+$ or $\times$. e.g. $M\odot v=Mv+(v^TMv)I$ is something that I remember as an example. The context of your book should clarify it though. $\endgroup$ – Daryl May 30 '14 at 20:28
  • $\begingroup$ here is an exampe, at 46. min youtu.be/iX5V1WpxxkY definition of LSTM $\endgroup$ – jhegedus Sep 1 '16 at 19:49
  • $\begingroup$ The Sun, maybe? en.wikipedia.org/wiki/Solar_symbol $\endgroup$ – Oscar Lanzi Oct 27 '16 at 1:09

In the LSTM equations, the circled dot operator is typically used to represent element-wise multiplication.

  • $\begingroup$ Is it possible to reconcile this interpretation with the OP's claim that "$M$ is a matrix and $N$ might be a vector, or a matrix, or a scalar"? $\endgroup$ – hardmath Sep 21 '16 at 1:33

When I researched about the symbol ⊙. I got two things:

$1$. In the book Meaning, Logic and Ludics-By Alain Lecomte, The writer says that: Let $U$ and $B$ be two positive designs we denote tensor product of $U$ and $B$ by $U$⊙$B$.

$2$. In the book Dag Prawitz on Proofs and Meaning: The writer says that $\alpha⊙\beta$ means that either $\beta $ is derivable from $\alpha$ or is $\alpha$ derivable from $\beta $.

I think that second on e makes sense, as for tensor product I have always seen the symbol ⊗ being used. I provided you the info which I had, hope it helps.

  • $\begingroup$ This is the correct answer! $\endgroup$ – Yahya Nov 7 '19 at 10:43

We use the circle with a dot notation in nonlinear continuum mechanics. It is defined in index notation as \begin{align} \left(\boldsymbol{M} ⊙ \boldsymbol{N}\right)_{ABCD}= \dfrac{1}{2}\left(M_{AC}N_{BD}+M_{AD}N_{BC}\right) \end{align} where $\boldsymbol{M}$ and $\boldsymbol{N}$ are second order tensors. The result of this operation is a 4th order tensor. You can think of it as related to the outer product of two matrices or $\boldsymbol{M}\otimes \boldsymbol{N}$ in that it takes two 2nd order tensors and generates a 4th order tensor. Remember the outer product is defined for two second order tensors as

\begin{align} \left(\boldsymbol{M} \otimes \boldsymbol{N}\right)_{ABCD}= M_{AB}N_{CD} \end{align}

The circle with a dot operation $⊙$ occurs when calculating the elastic modulus $\mathbf{\underline{C}}$ (a 4th order tensor) from constitutive laws. Examples include the Mooney-Rivlin or Neohookean material models. To get the elastic modulus you must take derivatives of tensors with respect to other tensors. For instance,

\begin{align} \underline{\mathbf{C}} = 2\dfrac{d\boldsymbol{S}}{d\boldsymbol{C}} \end{align} or in index notation as \begin{align} \text{C}_{ABCD} = 2\dfrac{\partial S_{AB}}{\partial C_{CD}} \end{align} where $\boldsymbol{C}$ is the left Cauchy-Green deformation tensor and $\boldsymbol{S}$ is the 2nd Piola-Kirchhoff stress tensor. Based on the constitutive models, which relate stress in a material to the strain, it turns out $\boldsymbol{S}$ is a function of the inverse of $\boldsymbol{C}$. If you work it out you will find that you will have to compute $\frac{\partial \boldsymbol{C}^{-1}}{\partial \boldsymbol{C}}$. The proof is a headache but the result comes out to \begin{align} \dfrac{\partial \boldsymbol{C}^{-1}}{\partial \boldsymbol{C}} = -\boldsymbol{C}^{-1} ⊙ \boldsymbol{C}^{-1} \end{align} In index notation \begin{align} \dfrac{\partial C^{-1}_{AB}}{\partial C_{CD}} = -\dfrac{1}{2}\left(C^{-1}_{AC}C^{-1}_{BD}+C^{-1}_{AD}C^{-1}_{BC}\right)=-\left(\boldsymbol{C}^{-1} ⊙ \boldsymbol{C}^{-1}\right)_{ABCD} \end{align} The circle with a dot operation only arises because $\boldsymbol{C}$ is a symmetric matrix, i.e., $\boldsymbol{C}=\boldsymbol{C}^{T}$ and $\boldsymbol{C}_{sym}=\dfrac{1}{2}\left(\boldsymbol{C}+\boldsymbol{C}^{T}\right) = \boldsymbol{C}$. Note that if taking the derivative of an inverse of a nonsymmetric tensor with respect to itself yields \begin{align} \dfrac{\partial A_{AB}^{-1}}{\partial A_{CD}}=-A^{-1}_{AC}A^{-1}_{DB} \end{align} and this is not the outer product. This operation has not yet been given a symbol.


  • the outer product $\otimes$ is also called the tensor product.
  • The indices are capital letters ABCD since in continuum mechanics capital letters denote Lagrangian/material (or reference configuration) coordinates. Lower case indices ijkl denote spatial/Eulerian coordinates.

Addendum: Other uses I have seen for $⊙$ include

  1. In physics, I have seen it mean a point source such as a point charge or gravity source like a planet.
  2. In physics, I have seen it mean the vector points out of the page $⊙$. And $\otimes$ means the direction of the vector is into the page. I have seen this in E&M for B-fields and E-fields and mechanics for torques.
  3. In mathematics it could mean a function composition operator, which maps functions to functions, e.g., $\,f⊙g$.

This is what I think it is used for in that Long Short-Term Memory neural networking lecture https://www.youtube.com/watch?v=iX5V1WpxxkY&feature=youtu.be at 46:00.

In mathematics, functional compositions are usually denoted by a small circle $\circ$. For example, Eulerian $f(\boldsymbol{x},t)$ and Lagrangian $F(\boldsymbol{X},t)$ descriptions are related to each other by a function composition: \begin{align} F(\boldsymbol{X},t)=f(\boldsymbol{\Phi}(\boldsymbol{X},t),t) \textrm{ or } F = f \circ \boldsymbol{\Phi} \end{align}


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