Summation Convention for 2 Vectors

[BHL 20]

From an exercise set on the summation convention: X and Y are given as [Xⁱ] = \begin{pmatrix}
1\\ 0\\ 0\\ 1\end{pmatrix} and [Yⁱ] = \begin{pmatrix} 0\\ 1\\ 1\\ 1\end{pmatrix} There are a few questions involving these vectors. The one I am stuck on asks to compute XⁱY^j . It may be necessary to raise lower indices in the question, the book that this question comes from uses a metric with signature ( - + + + ) for doing this.
I have no attempt, I have no idea what the question actually wants. I thought there is only a summation if the indices are the same and matrix multiplication is obviously not an option

 

[pasmith]

From an exercise set on the summation convention: X and Y are given as [Xⁱ] = \begin{pmatrix}
1\\ 0\\ 0\\ 1\end{pmatrix} and [Yⁱ] = \begin{pmatrix} 0\\ 1\\ 1\\ 1\end{pmatrix} There are a few questions involving these vectors. The one I am stuck on asks to compute XⁱY^j .

x^iy^j is the tensor of rank 2 (square matrix) which has x^i y^j at row i, column j.

 

[BHL 20]

So in other words it's a YX^T operation that's required. I am quite confused by your answer, is it supposed to be evident from the notation that this is the thing to do?

 

[Mark44]

BHL 20, please don't adorn your text with size tags.

 

[Fredrik]

So in other words it's a YX^T operation that's required. I am quite confused by your answer, is it supposed to be evident from the notation that this is the thing to do?

Maybe you should post the exact problem statement. "Compute $X^i Y^j$" doesn't make much sense to me. Does it mean "compute $X^iY^j$ for all i,j? Then the straightforward way is to just do these multiplications one at a time: $X^0Y^0=1\cdot 0=0$, $X^1Y^0=0\cdot 0=0$,... The results of these 16 calculations can be arranged in a matrix, which for all i,j, has $X^iY^j$ on row i, column j (and the results can be arranged in other ways as well), but you shouldn't be required to present the result in the form of a matrix.

If you really want to, you can use the definition of matrix multiplication in the following way. For all i,j, we have
$$(XY^T)^i{}_j =\sum_{k=0}^0 X^i{}_k (Y^T)^k{}_j = X^i{}_0 Y^j{}_0 =X^i Y^j.$$

 

[BHL 20]

Thanks Fredrik, I see now.