05

What a vector cannot say

A representation is useful because it preserves selected relationships. It is limited for the same reason.

A high score is not a verdict

“The charger is compatible” and “The charger is not compatible” discuss almost the same topic but make opposite claims. We have not measured either sentence's embedding here. The point is about the task: related subject matter is not the same as agreement, and agreement is not the same as truth.

A good retrieval evaluation therefore needs representative questions and judgments about which passages actually help. Geometry can generate candidates; the returned text and its provenance remain essential evidence.

A picture can hide a difference

Our lab is genuinely two-dimensional. To draw a higher-dimensional vector as a point in a 2D plot, we need a mapping to two coordinates. Consider the deliberately simple projection that drops the third coordinate:

FIG 005 A flat view can hide a difference
A flat view can hide a differencePoints [1,0,0] and [1,0,4] are four units apart. Dropping the third coordinate maps both to [1,0], so their displayed separation is zero.ORIGINAL: 3 COORDINATESa = [1, 0, 0]b = [1, 0, 4]distance = 4drop zsame 2D pointdisplay distance = 0A flat view can hide a differencePoints [1,0,0] and [1,0,4] are four units apart. Dropping the third coordinate maps both to [1,0], so their displayed separation is zero.ORIGINAL: 3 COORDINATESa = [1, 0, 0]b = [1, 0, 4]distance = 4Drop z. Both become [1, 0].same 2D pointdisplay distance = 0

Constructed dimension-drop example—not a projection of real embedding data. A picture can hide separation that still exists in the original space.

The original points [1,0,0][1,0,0] and [1,0,4][1,0,4] are 4 units apart. Both become [1,0][1,0] on the page, so their displayed distance is zero. This counterexample does not condemn every projection. It shows why a 2D picture is not automatically evidence about original-space distances.

Try a fresh pair

Take q=[1,0]q=[1,0], U=[0.6,0.8]U=[0.6,0.8], and V=[3,4]V=[3,4]. Which candidate wins under cosine, dot product, and Euclidean distance? What changes if you normalize every vector first?

Work through the answer

UU has norm 1 and VV has norm 5. They point in the same direction, since V=5UV=5U. Both cosine scores are 0.6: a tie. Their dot products are 0.6 and 3, so dot product prefers V.

∥q−U∥2=0.8≈0.8944∥q−V∥2=20≈4.4721\begin{aligned}\|q-U\|_2&=\sqrt{0.8}\approx0.8944\\ \|q-V\|_2&=\sqrt{20}\approx4.4721\end{aligned}

Euclidean distance prefers U. After normalization, V becomes U; q and U already have unit length. All three rules then tie the two candidates. None of these numbers tells you whether either passage is true.

Take the map—not the illusion

Keep three questions: What was this representation trained to preserve? Which comparison rule does it expect? Does the retrieved source actually help?

An embedding gives you coordinates for a task. The coordinates make comparison possible; the task, scoring rule, and evidence make the comparison meaningful.