While reviewing the Transformer architecture, I realized something I didn't expect, which is that :

  • the positional encoding is summed to the word embeddings
  • rather than concatenated to it.

positional encoding summed to word embedding


Based on the graphs I have seen wrt what the encoding looks like, that means that :

  • the first few bits of the embedding are completely unusable by the network because the position encoding will distort them a lot,
  • while there is also a large amount of positions in the embedding that are only slightly affected by the positional encoding (when you move further towards the end).

graph shows positional encoding affects firsts logits a lot, last logits hardly not


So, why not instead have smaller word embeddings (reduce memory usage) and a smaller positional encoding retaining only the most important bits of the encoding, and instead of summing the positional encoding of words keep it concatenated to word embeddings?

  • $\begingroup$ I was also curious about this, have you figured it out? $\endgroup$
    – Lee MJ
    Feb 1, 2020 at 14:34
  • $\begingroup$ @LeeMJ: No, I did not. $\endgroup$ Feb 4, 2020 at 11:26
  • $\begingroup$ Have you figured it out now? $\endgroup$ May 25, 2020 at 18:16
  • $\begingroup$ Is anyone aware of any papers where they tried concatenation instead of adding? $\endgroup$ Feb 24, 2021 at 16:29
  • $\begingroup$ @keith-johnson Not per se, but Google T5 does use a different approach, where position is encoded separately. Since there is a lot about Google T5 you can maybe also check this other paper that builds on top of T5 and tweaks its positional encoding some more: arxiv.org/pdf/2102.09550.pdf $\endgroup$ Feb 24, 2021 at 19:48

6 Answers 6


When you concatenate, you have to define a priori the size of each vector to be concatenated. This means that, if we were to concatenate the token embedding and the positional embedding, we would have to define two dimensionalities, $d_t$ for the token and $d_p$ for the position, with the total dimensionality $d = d_t + d_p$, so $d>d_t$ and $d>d_p$. We would be decreasing the total size we devote to tokens in favor of positional information.

However, adding them together is potentially a super case of the concatenation: imagine that there is an ideal split of $d$ into $d_t$ and $d_p$ in terms of minimizing the loss; then, the training could converge to position vectors that only take $d_t$ elements, making the rest zero, and the positions were learned and happened the same, taking the complementary $d_p$ elements and leaving the rest to zero.

Therefore, by adding them, we leave the optimization of the use of the $d$ dimensions to the optimization process, instead of assuming there is an optimal partition of the vector components and setting a new hyperparameter to tune. Also, the use of the vector space is not restricted by a hard split in the vector components, but takes the whole representation space.

  • 3
    $\begingroup$ This would make sense for learned positional encoding. What about the sine/cosine encoding? Does it just rely on the fact that nothing much is happening in dimensions beyond the first few? $\endgroup$
    – max
    Feb 24, 2021 at 5:56
  • 1
    $\begingroup$ While the equivalence of concatenation and addition may only apply to learned positional encoding, I think that the general optimization of the representation space does apply to fixed encodings as well (although the optimization only happens in the token embeddings). I don't think the picture is correct (it has changed in the referenced tutorial) $\endgroup$
    – noe
    Feb 24, 2021 at 8:24
  • 4
    $\begingroup$ maybe a stupid question, but why this addition doesn't spoil the embedding - like we had word king, add this pattern and receive slave ? $\endgroup$ Feb 25, 2021 at 21:39
  • 1
    $\begingroup$ If such a thing would happen, the final loss would be bad. The training aims at improving the loss, and therefore prevents that situation from happening. $\endgroup$
    – noe
    Feb 25, 2021 at 21:45
  • $\begingroup$ You mentioned adding them, concatenating them. How about other ways like vector-matrix multiplication? It's a linear transformation. Theroretically, can this be used to integrate positional information as well? Is it a bad approach because of too many parameters (one matrix per position)? $\endgroup$ Dec 3, 2022 at 9:48

the first few bits of the embedding are completely unusable by the network because the position encoding will distort them a lot

This confused me very much at first because I was thinking of the model using a pre-trained word embedding. And then an arbitrary initial chunk of that embedding gets severely tampered with by the positional encoding.

However, in the original transformer model at least, the embedding was trained from scratch, so this does not apply. An initial chunk of the overall embedding will be used for positional information, and the rest will be used for word information.

This still doesn't explain why we use this method instead of concatenation -- see the other answers for that -- but it does explain why the method isn't crazy.

That said, it may be that the method works well even with pre-trained word embeddings, I don't know. If so, it's hard to explain.


The confusion here is that we believe positional embedding is a more complicated version of adding positional information to the word embedding; however, it is not actually. Adding new dimensions to each embedding increases the dimensionality of the problem. On the other hand, please note that the added positional embedding is (almost) static, as shown in this image for a 2D positional embedding:

enter image description here

The added positional embeddings are the same for all the inputs, and the transformer can separate the positional information from the actual word embedding through the training process. Therefore, the positional embedding doesn't mess with the word embedding information, and adding them is a more efficient way of adding the positional information that concatenates them.

  • $\begingroup$ it is possible that embed("foo") != embed("bar"), however, embed("foo") + p1 == embed("bar") + p2. Thought, the embedding is learned, the foo and bar will not appear in all possible position, thus the network has no way to separate them when this happens occasionally. $\endgroup$
    – Wang
    Aug 3 at 15:42

It is been a while, but I think anyone ending up here might also be interested in the reading of the following paper:

What Do Position Embeddings Learn? An Empirical Study of Pre-Trained Language Model Positional Encoding (Yu-An Wang, Yun-Nung Chen)


I am not changing the accepted answer as this article is not specific.


The following is conjecture, not fact.

If you look at how much each scalar in the the positional embedding vector changes as a function of position... you'll find that many of the scalars barely change at all. You can visualize this with any positional embedding plot, where the x axis is usually the [512] length of the vector, and the y axis is the position of the token.

For example, this image is from Jay Alammar's well regarded "The Illustrated Transformer"

enter image description here

Let's try to do this mathematically as well. The implementation of PE's that Jay references is at this Google GitHub repo:


Running the function on a PE/WE of length 512 and max sentence length of 128, let's look at how much the final value in the vector actually changes from the first position, to the 64th position, to the final position. Answer: not much.

print(signal[0, 0, -1])
print(signal[0, 63, -1])
print(signal[0, 127, -1])

tf.Tensor(1.0, shape=(), dtype=float32)
tf.Tensor(0.99998015, shape=(), dtype=float32)
tf.Tensor(0.99991935, shape=(), dtype=float32)

Ditto for a value 16 steps away from the final location:

print(signal[0, 0, -16])
print(signal[0, 63, -16])
print(signal[0, 127, -16])

tf.Tensor(1.0, shape=(), dtype=float32)
tf.Tensor(0.9984067, shape=(), dtype=float32)
tf.Tensor(0.9935305, shape=(), dtype=float32)

I saw elsewhere that BERT's WEs are typically roughly the range of [-2, 2], so adding a 0.007 delta from the PE would not move the WE very much at the -16th position.

So what I think is probably happening is that only ~256 of the PE vector's values are actually moving around as a function of the position... the rest are ~constant. Then the learned WE (Transformers don't use prelearned WE like word2vec or glove), figures out to only use the other ~256 elements. So really... it's conceptually a concat.

notebook here


  • $\begingroup$ Thank you for the interesting analysis. I would note that, however, many transformers keep the positional embeddings trainable, so while the initial value gives an a-priori for the transformer to use the last positions only, the accepted answer is correct that the model can learn to use more for the words by zeroing positions it doesn't find useful. $\endgroup$ Feb 9 at 14:29
  • 1
    $\begingroup$ Totally agree. The "Attention is All You Need" paper said they tried trainable and fixed PE, and got similar results, but more recent transformers like ViT do train the PE. If both the PE and WE are trainable, then sum is okay, since the PE and WE path can adaptively learn to coexist - conceptually, they can dynamically learn how many elements the PE gets and how many the WE gets. Not unlike a residual skip connection. But if the PE were fixed and very noisy across all elements, the intuition most people have is that it'd be impossible to train a WE or use pretrained WE like word2vec. $\endgroup$
    – Yaoshiang
    Feb 9 at 20:45
  • $\begingroup$ if this is true, we can basically use half of the embedding vector length then concatenate the half width embedding and positional vector to get exactly same final matrix. $\endgroup$
    – Wang
    Aug 3 at 15:48
  • $\begingroup$ @Wang, I agree with that directionally. A key would be to understand how to divide up the embedding - doesn't have to be 50/50. Maybe 80/20 or 95/5 ends up being a better use of the vector. A trainable PE would probably be nearly ideal in using the right amount of information. $\endgroup$
    – Yaoshiang
    Aug 25 at 20:31

So the question is about why positional embeddings are directly added to word embeddings instead of concatenated. This is a particularly interesting question. To answer this question, I will need to firstly separate the differences between sequential networks like RNNs and Transformers, which then introduces this problem nicely.

In RNNs, we feed in data (let's say a sequence of words) into the model in a sequential manner. This means that in the context of inputting in a sequence of words, the model does arguably obtain the order the tokens as it is fed in one by one.

With transformers, on the other hand, all of the words in the sequence are fed in all at once. This means that, so far, transformers do not have any notion of word ordering. Therefore, we need positional embeddings to tell the model where each word belongs in the sequence.

I believe the reason why we add them to word embeddings is because we want to maintain a similar input into the model as an RNN, which takes in word embeddings as its input as well. I think your question is a very good one to ask, and maybe you should experiment with having a more compressed word embedding with its positional embedding and compare your approach against the more "traditional" approach and see what results you yield. I'll be excited to see them.


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