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I'm trying to understand a paper from Brakerski and Vaikuntanathan. In it, they use the notation $\langle a,b \rangle$ but they don't explain what that means.

The answers on this question say that it's just another way to write a tuple, but I don't think that's right in this case since the authors use the regular $(a,b)$ notation in the rest of the paper to denote a tuple.

Does anyone know what else $\langle a,b \rangle$ could mean?

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    $\begingroup$ It's inner product. $\endgroup$ – conchild Jan 21 at 16:13
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    $\begingroup$ See the top of page 9 of that paper. It is the inner product. $\endgroup$ – Yehuda Lindell Jan 21 at 16:48
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In mathematics, $\langle a,b \rangle$ represents the inner product of two vectors and this is a generalization of the dot product. It is one of the multiplications of the vectors in a Vector space and the result is a scalar.

An inner product satisfies these properties

  1. $\langle c+a,b\rangle =\langle c,a\rangle +\langle a,b\rangle $.

  2. $\langle \alpha a,b\rangle = \alpha\langle a,b\rangle$ where $\alpha$ is a scalar.

  3. $\langle a,b\rangle =\langle b,a\rangle$.

  4. $\langle a,a\rangle = 0 $ and equal if and only if $v=0$.

    • The fourth condition is the positive definite condition. Some authors define the inner product with only the first 3 conditions with a weaker 4th condition $\langle a, b \rangle =0$ for all $b$ then $a=0$.

    As noted in the comments by @levgeni, if the field is $\mathbb{F}_2$ then

    $$\langle (1,1),(1,1) \rangle = 1 \cdot 1 + 1 \cdot 1 = 2 = 0$$ and $$\langle (1,1),(1,0) \rangle = 1 \cdot 1 + 1 \cdot 0 = 1 = 1$$ therefore, the weaker condition is required.

The inner product definition changes according to the vector space. In the context of the paper, let

$$a =(a_1,\ldots,a_n)$$ $$b =(b_1,\ldots,b_n)$$

be two n-dimensional vectors then;

$$\langle a, b \rangle = a_1 \cdot b_1 + \cdots + a_n \cdot b_n $$

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    $\begingroup$ In a Cryptographic context, the fourth condition is not verified; if the field is $\mathbb{Z}_2$, $\langle \left(1,1\right), \left(1,1\right) \rangle = 1\cdot1 + 1\cdot1 = 0$ $\endgroup$ – Ievgeni Jan 22 at 13:53
  • $\begingroup$ @levgeni Better now? $\endgroup$ – kelalaka Jan 22 at 14:24
  • $\begingroup$ I don't see anything to add. It seems okay to me. $\endgroup$ – Ievgeni Jan 22 at 14:36

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