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Algebra isomorphism

  1. Mar 5, 2012 #1

    quasar987

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    Consider the complex numbers C as an algebra over the reals R. The author of the book I have in front of me (Dirac operators in Riemannian Geometry, p.13) writes

    [tex]\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}=\mathbb{C}\oplus\mathbb{C}[/tex]

    (as real algebras). Does anyone know what this canonical algebra isomorphism is??? Obviously, woz -->(w,z) is not even linear.
     
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  3. Mar 5, 2012 #2

    micromass

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    Note that this actually follows from

    [tex](A\oplus B)\otimes C \cong (A\otimes C)\oplus (B\otimes C)[/tex]

    Thus

    [tex]
    \begin{eqnarray*}
    \mathbb{C}\otimes_\mathbb{R}\mathbb{C}
    & \cong & (\mathbb{R}\oplus \mathbb{R})\otimes_\mathbb{R} \mathbb{C}\\
    & \cong & (\mathbb{R}\otimes_\mathbb{R} \mathbb{C})\oplus (\mathbb{R}\otimes_\mathbb{R}\mathbb{C})\\
    & \cong & \mathbb{C}\oplus \mathbb{C}\\
    \end{eqnarray*}
    [/tex]

    This first isomorphism sends [itex](x+iy)\otimes z[/itex] to [itex](x,y)\otimes z[/itex].
    The second isomorphism sends [itex](x,y)\otimes z[/itex] to [itex](x\otimes z, y\otimes z)[/itex].
    The third isomorphism sends [itex](x\otimes z, y\otimes z)[/itex] to [itex](xz,yz)[/itex]

    So your total isomorphism sends [itex]x+iy\otimes z[/itex] to (xz,yz).
     
    Last edited: Mar 5, 2012
  4. Mar 5, 2012 #3

    quasar987

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    I came to consider this vector space isomorphism at some point, but realized that it is not an algebra isomorphism:

    [tex]((x_1+iy_1)\otimes z_1)((x_2+iy_2)\otimes z_2)=((x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)\otimes (z_1z_2)\mapsto ((x_1x_2-y_1y_2)z_1z_2,(x_1y_2+x_2y_1)z_1z_2)[/tex]

    while

    [tex](x_1z_1,y_1z_1)(x_2z_2,y_2z_2)=(x_1x_2z_1z_2,y_1y_2z_1z_2)[/tex]
     
  5. Mar 5, 2012 #4

    morphism

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    As a ring, $$ \mathbb C = \mathbb R[x]/(x^2+1) $$ so $$ \mathbb C \otimes_{\mathbb R} \mathbb C = \mathbb C[x]/(x^2+1) = \mathbb C[x]/(x+i) \oplus \mathbb C[x]/(x-i) = \mathbb C \oplus \mathbb C.$$
    If you follow the above maps, you get the following explicit ##\mathbb R##-bilinear map $$ \begin{align} \mathbb C \oplus \mathbb C &\to \mathbb C \oplus \mathbb C \\
    (z_1,z_2) &\mapsto (z_1z_2, z_1 \overline{z_2}) \end{align}$$ which implements the above isomorphism ##\mathbb C\otimes_{\mathbb R}\mathbb C \stackrel{\sim}{\to} \mathbb C \oplus \mathbb C##.
     
  6. Mar 5, 2012 #5

    quasar987

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    hail morphism! :)
     
  7. Mar 6, 2012 #6

    mathwonk

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    that is very nice. another way to look at morphism's map is to note that the tensor product is the "categorical sum" of R algebras, hence to define an algebra map out of the tensor product is equivalent to defining one algebra map out of each factor.

    Moreover an algebra map should send the multiplicative identity to the multiplicative identity, so 1tens1 should go to (1,1).

    Then guided by his map one notes that the two algebra maps z-->(z,z) and w-->(w,wbar),

    must induce an algebra map from the tensor product (categorical direct sum), to the module direct sum,
    sending ztens1 to (z,z) and (1tensw) to (w,wbar), hence (ztensw) to (zw,zwbar).

    This is easily checked to be injective as vector space map, by checking the R- independence of the images

    of the vector basis 1tens1, 1tensi, itens1, itensi.

    hence an isomorphism of algebras.

    I don't see the inverse map so easily, help???
     
  8. Mar 7, 2012 #7

    mathwonk

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    More general mumbo jumbo:

    When you want to show two things are isomorphic, the first job is to find a map between them and then hope it is an isomorphism.

    It is easy to define a map out of a categorical "direct sum", or lets call it coproduct, since in this case the algebra that looks like a direct sum is not a categorical sum.

    So any pair of maps, one out of each object, defines a unique map out of the "coproduct", in this case the tensor product of algebras.

    On the other hand a categorical "product" has the opposite property, namely it is easy to define maps into it. Any pair of maps, one into each factor, defines a unique map into the "product". In our case it appears to me as if the direct sum algebra above, is actually a categorical product. That is why it is easier to define a map from the tensor algebra into the direct sum algebra than the other way. I.e. it is much easier to define a map from a coproduct to a product than the other way around.

    I pointed out above that we get morphism's map out of the tensor algebra from the pair of maps z-->(z,z), and w-->(w,wbar), out of C.

    But we could also get the same map from the pair of maps ztensw-->(zw), and ztensw-->(zwbar), both into C.

    Of course this construction also used the "sum" property of the tensor product to define the two maps. But these two maps are about the only ones one could think of. Certainly ztensw-->zw is canonical, and then we get the other by changing it by the only automorphism of C over R, namely conjugation.

    The only other map I could think of would be ztensw-->zbar.wbar, which does not give an injection when paired with ztensw-->zw, since 1.tens.i and i.tens.1 have the same image.

    Presumably however the pair ztensw-->zbar.wbar, and ztensw-->zwbar,

    would also work. I.e,. we could also map ztensw --> (zbar.wbar, z.wbar), as this is essentially morphism's map followed by an automorphism.

    I still don't see the map back.
     
    Last edited: Mar 7, 2012
  9. Mar 7, 2012 #8

    mathwonk

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    Oh I guess morphism's original point of view makes it clear how to write down the inverse. I.e. he shows the isomorphism is essentially given by the "chinese remainder theorem".

    So if one can write down an inverse to the map Z/(mn)-->Z/m x Z/n, where gcd(m,n) = 1, then one should be able to write down an inverse to:

    C[x]/(x^2+1)=C[x]/(x+i)⊕C[x]/(x−i).
     
  10. Mar 7, 2012 #9

    morphism

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    Let's call the map going out of the tensor f and its inverse g. To determine g, it's enough to determine g(1,0) and g(0,1). It's easy to see that these aren't going to be elementary tensors (e.g. if ##g(1,0)=z\otimes w## then ##(zw,z\bar{w})=(1,0)## which is nonsense). The next best thing we could hope for is a sum of two elementary tensors. After some playing around, I managed to find that
    $$ g(1,0)=\frac{1}{2}(1\otimes 1 - i\otimes i) \quad \text{ and } \quad g(0,1)=\frac{1}{2}(1\otimes1 + i\otimes i). $$ Unfortunately I don't have a "conceptual" explanation here.

    Part of the difficulty is that we're essentially trying to invert the evaluation maps ##\mathbb C [x]/(x\pm i) \to \mathbb C##, which isn't very natural.
     
  11. Mar 7, 2012 #10

    mathwonk

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    That seems to agree with what I got, namely

    (u,v) goes to (1/2)tens(u+vbar) - (i/2)tens(u-vbar)i.

    The idea I used (of yours) was that since the map from C[x] to CxC which induces the isomorphism, takes a polynomial to its pair of values at i and -i, to invert it we are trying, as you say, to find a polynomial in C[x]/(x^2+1) whose value at i and -i gives us u and v. But there is a standard trick for this, using euclid's algorithm for gcd's.

    I.e. to invert the chinese remainder map Z/(mn)-->Z/m x Z/n we solve an+bm = 1, and then an goes to (1,0) and bm goes to (0,1).

    Since (x-i) - (x+i) = 2/i, we get (i/2) (x-i) - (i/2) (x+i) = 1, hence the

    polynomials (i/2)(x-i) and (-i/2)(x+i) having the values (0,1) and (1,0) at i and -i.

    But this was apparently off by a conjugate, so I fudged it at the end, adding bar to the v's.
     
    Last edited: Mar 7, 2012
  12. Mar 7, 2012 #11

    mathwonk

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    I guess there is an even simpler way to find a polynomial which equals 1 at i and 0 at -i, by starting from any polynomial P which has different values at the two points, and forming

    {P(x) - P(-i)}/{P(i) - P(-i)}. E.g. we could start with x+i, ending up with (x+i)/2i.

    This is no doubt a very special case of Lagrange interpolation, which inverts the evaluation problem in general. Since the polynomial is unique if it has degree less than the number of points, this works well for us in C[x]/(x^2+1). We get a unique linear polynomial.

    Thank you morphism, for teaching us how to solve this exercise that I had seen long ago, but never worked!
     
  13. Mar 7, 2012 #12

    mathwonk

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    no doubt morphism's playing around was essentially solving lagrange's interpolation problem in this case.

    i.e. to find a polynomial of degree at most two with any desired values at say a,b,c, all different, just start with (x-b)(x-c)/(a-b)(a-c). that has value 1 at a, and zero at b,c. Then do the same for b and c, and take a linear combination to get the values you want.

    So to find a linear polynomial with any desired values at i and -i, start with

    (X+i)/(i+i)) and (X-i)/(-i-i). These give 1 and 0, or 0 and 1, at i and -i.


    This function point of view is actually the same as the euclidean algorithm method for inverting the chinese remainder map. I.e. to find an integer that maps to (1,0) in Z/n x Z/m, we start with an integer that maps to 0 in Z/m, such as m. That is like starting from the polynomial X-i that evaluates to zero at i.

    Then just as we divided X-i by its value at -i, to make the value 1, so also we want to divide m by its value mod n. I.e. we want to multiply by the inverse of m mod n. But if an +bm = 1, then the inverse of m is b, mod n. So we modify our answer from m to bm. Then bm goes to (1,0) in Z/n x Z/m. (I.e. since i/2 is inverse to -2i, hence bm is analogous to (i/2)(X+i).)
     
    Last edited: Mar 7, 2012
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