# Inversion w.r.t. a sphere: Operator

1. Jun 9, 2012

### neelakash

Hello everyone,
I have enquired about inversion in a sphere here in past: https://www.physicsforums.com/showthread.php?t=440759

Although that time I could not come back to the discussion (apologies to Jason), later I went through some of the properties mentioned by him. This opened a wider scope applications in front of me where inversion could be of use. However, I never found the answer to my present question: is it possible to obtain an explicit operator form of the inversion w.r.t. a sphere?

If we get such an operator, life will be very easy for a physicist. We just subject a given object to the operator and we will invert it w.r.t. a sphere. This may be a helpful way for the understanding of image problems.

Can anyone tell me how to obtain the operator given the rule of coordinate transformations is
$$(r,\theta,\phi)\rightarrow(\frac{a^2}{r},\theta,\phi)$$?
Can the fact that this is a conformal map help to obtain the operator?

-Neel

Last edited: Jun 9, 2012
2. Jun 9, 2012

### Muphrid

If by "operator" you mean a linear operator along the lines of rotations or inversion through the origin, I think it's pretty clear that this transformation is not linear, so that wouldn't happen in a traditional 3D Euclidean space framework.

However, if you investigate conformal geometric algebra, I'm fairly certain that inversions with respect to spheres are linear operators in that framework. Regardless, even with a nonlinear transformation, you can do lots of handy things with the Jacobian that underlies the transformation.

3. Jun 9, 2012

### neelakash

Hi, thank you for your reply...I was talking about inversion w.r.t. the sphere only...If by conformal geometry methods, it is possible to obtain linear inversion operator, that is a great news

I have started reading Needham's Visual Complex Analysis (3rd ch). I see he is giving a formula for inversion of a complex number z...Let me see if I can find out how to construct the operator...

4. Jun 9, 2012

### Vargo

Define "operator"...

5. Jun 10, 2012

### neelakash

By operator I mean a matrix, which when acts on a vector, transforms it to another. Typically a linear translation operator $$\hat{T}$$ acts like : $$\hat{T}_a [ (x) ]= x+a$$
-where a is the translation. Now, of course $$T$$ is a linear operator.

I was thinking about an analogous operator for inversion in sphere: say, $$I$$ which will have the following property: $$\hat{I}_a [ (r) ]= \frac{a^2}{r}$$ where a is the radius of the sphere...And it is supposed to keep $$\theta$$ and $$\phi$$ the same.

Do you get the idea?

-Neel

6. Jun 10, 2012

### mnb96

Did you try to have a look at the conformal model in geometric algebra? Try for example some sources by Hestenes, for example "spherical conformal geometry with geometric algebra" (section 6).
Sorry, I'm away and can't really elaborate more. Let me know if that was any use.

7. Jun 10, 2012

### neelakash

I have not seen it so far...I am downloading it...But I am getting up to ch4..not beyond that...

8. Jun 10, 2012

### Muphrid

The thing is, translations can't be expresssed as linear operators in plain Euclidean geometry. You can't add two vectors and apply the operator and get the same result as if you applied the operator to each and then added.

Linear operators must have a couple basic properties:

$$\underline A(b) + \underline A(c) = \underline A(b + c) \\ \underline A(\alpha b) = \alpha \underline A(b)$$

For vectors $b,c$ and scalar $\alpha$.

9. Jun 11, 2012

### neelakash

I guess you are right...I was thinking in terms of QM where we deal with linear operators...

In QM, a transformation in the real $ℝ^3$ induces an unitary transformation in the Hilbert space. Let's say, we have a given wave function $\psi(x)$ in a given reference frame(un-primed coordinates). We operate it with an unitary operator $\hat{U_O}$. Due to this operation, in Hilbert space, the wave function transforms as $\psi\rightarrow\psi'=\hat{U_O}[\psi]$, and in $ℝ^3$, $x\rightarrow\ x'$ where $x' =\hat{O}(x)$. Such a manipulation on the wave function (vector) is known as active transformation.

Since wave function is a scalar function, we must have $\psi'(x')=\psi(x)=\psi(\hat{O}^{-1}(x'))$...(note that this is an independent criterion)

Thus, $\psi'(x)=\hat{U_O}[\psi(x)]=\psi(\hat{O}^{-1}(x))$

As a concrete example, a translation operator is given as $x\rightarrow\ x'=\hat{O_a}(x)=x+a$ and the corresponding unitary operator is given as $U_a=e^{(-\frac{i}{\hbar}\ a\hat{\vec{p}})}$ where $\hat{\vec{p}}$ is the generator of the infinitesimal translation (momentum operator).

Now that I set the scene, let me restate the problem of inversion.

I am trying to find out the operator $\hat{I_a}$ that may correspond to the inversion on the surface of a sphere of radius $a$ in the real $ℝ^3$ space: $\vec{r}\rightarrow\vec{r'}=\hat{I_a}(\vec{r})=(\ a^2/r,\theta,\phi)^T$. Once it is found, I guess it will be possible to construct $\hat{U_I}$ so that we can obtain $\psi'(\vec{r})=\hat{U_I}[\psi(\vec{r})]=\psi(\hat{I_a}^{-1}(\vec{r}))$, if $\psi$ is known. However, whether this unitary operator can be constructed or not depends on whether the inner product remains conserved in these two spaces. We can't say anything about it now.

Please let me know if I could convey the problem...

Inversion in a sphere (in $ℝ^3$) is a conformal mapping (Inversion Theory and Conformal Mapping By David E. Blair; Ch-5, page 84, theorem 5.1)...I was thinking if this information can be used for this inversion process. Can we say that the operator $\hat{I_a}$ is related to the Jacobian matrix of the transformation: $(r,\theta,\phi)\rightarrow(\ a^2/r,\theta,\phi)$?

-Neel

Last edited: Jun 11, 2012
10. Jun 11, 2012

### Muphrid

Yeah, let me elaborate a bit on the theory of coordinate transformations in real vector spaces. We don't generally assume that the transformation from $r \to f(r) = r'$ (these should be understood as vectors, even though I won't dress them up) is linear, but you can do quite a bit with the Jacobian that underlies the transformation.

(Most of this is elaborated on, in this notation, in Doran and Lasenby's book on geometric algebra or their paper on Gauge Theory Gravity.)

Let a scalar field $\phi(r) = \phi'(r')$. Now, the chain rule can be used to relate derivatives of the two:

$$a \cdot \nabla \phi(r) = a \cdot \nabla \phi'(r') = [a \cdot \nabla f(r)] \cdot \nabla' \phi'(r') = a' \cdot \nabla' \phi'(r')$$

where $a' = a \cdot \nabla f(r) \equiv \underline f(a)$ defines the Jacobian. However, there's an even more powerful trick here:

$$a \cdot \nabla \phi = \underline f(a) \cdot \nabla' \phi' = a \cdot \overline f(\nabla') \phi'$$

Where $\overline f$ is the transpose or adjoint of the Jacobian (I'll usually call it the latter). From this, we can factor out the dot product with $a$ because it's an arbitrary vector, and we get

$\nabla = \overline f(\nabla')$

Similarly, consider a path $r(\tau)$ and its velocity $dr/d\tau$. These can be related to the $r'$ space by

$$\frac{dr'}{d\tau} = \frac{dr}{d\tau} \cdot \nabla r' = \underline f \left(\frac{dr}{d\tau}\right) \implies \underline f^{-1} \left(\frac{dr'}{d\tau} \right) = \frac{dr}{d\tau}$$

This is often used in, for example, transforming four-velocities in one frame to another in relativity. Indeed, relativists should realize that this means $\overline f$ acts on "covariant" vectors and $\underline f^{-1}$ acts on contravariant ones.

This is all the information needed to perform operations on coordinate transformations. One subtlety is that integrals need to be handled with care, because often we reduce integrands to scalars that really involve areas and volumes and would pick up factors of $\underline f^{-1}$ for each dimension. This can be handled elegantly in geometric algebra, but that's really a separate topic.

11. Jun 11, 2012

### Vargo

Here is a formula for the transformation in rectangular coordinates. Is this what you are looking for?

$$\vec{x} \mapsto r^2\frac{\vec{x}}{\|\vec{x}\|^2}$$

12. Jun 11, 2012

### neelakash

@Vargo: I am talking about the transformation you have written down. However, this is not all. I want to see the operator $\hat{I}$ that induces such a transformation.

@muphrid: I followed your calculation...Jacobian indeed proves to be a very powerful tool...

I found the Jacobian of this transformation in Blair's book. the Jacobian is a scalar times an orthogonal matrix.

Now there are few questions:

(a) Can we say the points on a radius of the sphere form a group under multiplication?

apparently, yes...closure is OK...existence of identity and inverse element is fine...associativity should pose no problem.

In fact, after last few days study, inversion in a sphere is a kind of Mobius transformation; so these should form a Mobius group...

(b) Given the jacobian, is it possible to obtain the generator of the group?

I do not know the answer of the question now...However, for a rotation by the operator $\hat{ℝ}(\theta)$, which induces $x\mapsto\ x'=\hat{ℝ}(\theta)(x)$, the Jacobian matrix is equal to the rotation matrix $\hat{ℝ}$ itself. And this rotation $\hat{ℝ}$ in $ℝ^3$ induces a unitary transformation $\hat{U_ℝ}$ in the Hilbert space; the latter can be exponentiated in the following form: $\hat{U_ℝ}(\theta)=e^{(i\vec{\theta}\cdot\hat{\vec{J}})}$ where $\hat{\vec{J}}$ is the generator of rotation, namely the angular momentum operator.

Here, in case of inversion, I have the Jacobian matrix, but I cannot understand the next step. Eventually, I would like to obtain something like: $\hat{U_I}=e^{(i\hat{G_I})}$...However, unlike rotation, inversion does not seem to be an orthogonal matrix. So, I do not know if it is possible to obtain a unitary matrix for inversion...

-Neel

13. Jun 11, 2012

### Muphrid

This isn't complex/Hilbert space; there's no need to talk about unitary operators. The space of operators on vectors is much simpler.

It's always true that, for an operator $\underline S$ and its adjoint $\overline S$ that $\underline S(a) \cdot b = a \cdot \overline S(b)$.

Symmetric operators obey the property that $\underline S = \overline S$. The operator is equal to its adjoint (transpose).

Orthogonal operators obey the property that $\overline S = \underline S^{-1}$. Rotations are orthogonal, for example.

The operator that is both symmetric and orthogonal ought to be what, the identity operator?

I'm confused as to what you mean by the points on a sphere forming a "group" under multiplication. What definition of multiplication do you use that allows you to multiply two vectors to get another vector on the sphere? Cross products are problematic because they're undefined outside of 3D, also, so for the general case, one would want to use an extension like the outer (wedge) product.

Of course, the points on a sphere can be rotated to any other point on the sphere. There is a group of rotation operators, of course, and it's a two parameter family based on the angles you wish to rotate by.

Now, the following will only make real sense with an understanding of geometric algebra, but for the moment, let's say that objects like planes are valid geometric objects just as vectors (or lines) are, and that you can talk about linear combinations of planes the same way you can talk about linear combinations of vectors. Finally, let's say that you can multiply planes with other planes or with vectors in a way that makes sense.

If you can accept all that, then you'll be interested to know that rotations in a given plane $i$ have the following form:

$$\underline R(a) = \exp(-i\theta/2) a \exp(i\theta/2)$$

This should look familiar from a QM perspective, in terms of sandwiching exponentials. I've called the plane $i$ because under multiplication with itself, if it has unit magnitude it squares to -1, just like an imaginary unit. With each plane in n-dimensional space, there is an associated rotation parameterized in this fashion, and vectors in that plane stay in the plane. The plane itself acts as a generator of sorts for rotations within that subspace.

Additionally, when the exponential is expanded out by power series, one gets,

$$\exp(-i \theta/2) = \cos \theta/2 - i \sin \theta/2$$

People familiar with quaternions should recognize this form. The only difference is, perhaps, a matter of sign convention and understanding how to multiply planes with vectors in a logical manner.

I'm not sure what it would mean to have a generator of a group and how the Jacobian could define such a group, though. Some things between Euclidean geometry and Hilbert spaces are the same, but others are not. Operators on function can be linear in ways that operators on vectors are not. I think it's important to remember thta in Euclidean vector spaces, it is the vectors that we must be considering linear transformations on, not functions of vectors, but perhaps someone with more expertise can shed light on the similarities and differences.

14. Jun 11, 2012

### neelakash

I think there is some communication gap here...I was not talking about the points on the sphere forming a group under multiplication...this is what I had written:
however, let me think more, for I don't think I can take simple arithmetic multiplication of the distances as the group law...

15. Jun 11, 2012

### Muphrid

Well, points on the radius obey certain properties, sure, but again, I ask: what do you mean by multiplication?

16. Jun 11, 2012

### neelakash

Well, the points $P_n$ are such that $OP_n\subset(0,\infty)$ where $O$ is the center...Then, my set members are all these numbers $OP_n$s. One can multiply them to get another number in this set...Identity and inverse exist. Associativity also works...so they form a group...

However, now I do not think this notion will work...

17. Jun 11, 2012

### neelakash

I am sorry I could not formulate the question properly; it was a mistake to enquire about the group property of all the points lying on a radius of a sphere. Since, I am interested in finding the inversion (in a sphere) operator, the question I should have asked is if the set of all inversions in a sphere form a group...

Last edited: Jun 11, 2012
18. Jun 11, 2012

### neelakash

I am trying to give an answer to the question; please rectify me if I am wrong...Consider points on the radius of a sphere of radius $a$. The inverse transformation is given as $\hat{I}(r)=\frac{a^2}{r}$...We are trying to see if the set of all inversions form a group.

Identity: $\hat{I}(a)=\frac{a^2}{a}=a$...thus identity element $\hat{I}_e$ is there.

Inverse: $\hat{I_1}(r_1)=\frac{a^2}{r_1}$. We can say inverse of $\hat{I_1}$ exists, say $\hat{I_2}$ if we can show $\hat{I_2}(\frac{a^2}{r_1})=r_1$; the latter is true, however, as $\hat{I_2}$ is also an inversion operator defined as above. Thus, $\hat{I}_2={\hat{I}_1}^{-1}$

It appears that the closure and associative rules also be satisifed, if we notice that there is only few members of the group: identity $\hat{I}_e$, $\hat{I}_1$ and its inverse ${\hat{I}_1}^{-1}$...

What do you say?

-Neel

19. Jun 11, 2012

### neelakash

In fact, it looks like a Lie group with $r$ being the continuous parameter. With continuous variation of $r$, apparently all the group elements can be generated starting from the identity...