Why Can't We Do Algebraic Methods with Tensors?

[cr7einstein]

Hello everyone!
Even though I have done substantial tensor calculus, I still don't get one thing. Probably I am being naive or even stupid here, but consider

$$R_{\mu\nu} = 0$$.
If I expand the Ricci tensor, I get
$$g^{\sigma\rho} R_{\sigma\mu\rho\nu} = 0$$.
Which, in normal algebra, should imply,
$$ g^{\sigma\rho} = 0$$ (which is meaningless) or $$R_{\sigma\mu\rho\nu} = 0$$ ( which isn't always true).

So, Why can't we do normal algebra here?( it is perfectly valid step in algebra)
Also, consider a simple case
$$dS^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$.
Here, why can't we simply transpose(or divide both sides by) the differentials on RHS, i.e.,
$$\frac{dS^2}{dx^{\mu}dx^{\nu}} = g_{\mu\nu}$$ ?
Why is this expression not valid? Or, another example, Why can't
$$R_{\mu\nu} = g^{\sigma\rho} R_{\sigma\mu\rho\nu}$$ imply that
$$g^{\sigma\rho} = \frac{R_{\mu\nu}}{R_{\sigma\mu\rho\nu}}$$ ??
Is there a reason why this is wrong? Or is there a different way to transpose tensors from one side of the equation to the other side? Can you do this to vacuum field equations(as an example)?
Thanks in advance!

 

[Nugatory]

All of these problematic expressions use the Einstein summation convention. Any time that you find yourself wondering about whether an algebraic manipulation is valid in such an expression, you can expand the expression.

For example, you ask why ##A^{\sigma}B_{\sigma\rho\mu}=0## doesn't necessarily imply that either ##A## or ##B## are zero. If you write the summation out (in two dimensions to keep things simple) you get ##A^0B_{0\rho\mu}+A^1B_{1\rho\mu}=0##, which can be true even if none of the components of A or B are zero.

 

[HallsofIvy]

Exactly what "algebraic methods" do you mean? For many algebraic structures, such as matrices, "AB= 0" does NOT imply "A= 0 or B= 0".

 

[sweet springs]

Hi. This is not more than #2. I assume your normal algebra means product of numbers like 2 X 3 = 6. Do you know inner product of vectors like [tex]\mathbf{a}\cdot\mathbf{b}=0
[/tex]? This means vector a and vector b is orthogonal. a or b does not have to be a zero vector. For example a=(1.0) and b=(0,1) satisfy the eauation. What you referred is inner product of vector and tensor. Vector or tensor does not have to be a zero vector or tensor as well.

 

[Fredrik]

$$g^{\sigma\rho} R_{\sigma\mu\rho\nu} = 0$$.
Which, in normal algebra, should imply,
$$ g^{\sigma\rho} = 0$$ (which is meaningless) or $$R_{\sigma\mu\rho\nu} = 0$$

What you have there is of the form ##\operatorname{Tr}(AB)=0##, where A and B are square matrices. This doesn't imply that one of the matrices must be zero.

Definition of matrix multiplication: ##(AB)_{ij}=A_{ik}B_{kj}##.
Definition of trace: ##\operatorname{Tr}A=A_{ii}##.
$$\operatorname{Tr}(AB)=(AB)_{ii}=A_{ik}B_{ki}.$$

$$dS^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$.
Here, why can't we simply transpose(or divide both sides by) the differentials on RHS, i.e.,
$$\frac{dS^2}{dx^{\mu}dx^{\nu}} = g_{\mu\nu}$$ ?

Mainly because of the summation. What you're doing there is like dividing both sides of ##z=ax+by## (where all variables represent real numbers) by one of ##xx,xy,yx,yy## and (incorrectly) ending up with either ##a## or ##b## on the right-hand side.